Chemistry · Chapter 4

Stoichiometry of Chemical Reactions

A chemical equation is a recipe written in formulas. It names what goes in, what comes out, and in what ratio. This chapter turns that recipe into numbers you can trust. First you will balance an equation so that no atom appears or vanishes. Then you will sort reactions into a few families, so a new one looks familiar. You will track oxidation numbers to see which atoms gave electrons away. Last you will use the balanced equation as a bridge. Grams become moles, moles of one substance become moles of another, and moles become grams again. By the end you can predict how much product a real run should give.

By the end of this chapter you can

Balance a chemical equation by counting atoms on both sides.
Write the molecular, complete ionic and net ionic forms of a reaction in water.
Use solubility rules to predict whether a solid falls out of a mixture.
Spot the acid and base family and the redox family from an equation.
Assign oxidation numbers and name what was oxidized and what was reduced.
Convert grams to moles to moles to grams with a mole ratio.
Find the reactant that runs out first, then report theoretical and percent yield.

Section 1Writing and balancing equations

What a chemical equation says

A chemical equation is a short written record of a chemical change. The starting substances, called the reactants, sit on the left side. The substances that form, called the products, sit on the right side. An arrow joins the two sides, and that arrow means turns into. Here is methane burning in air.

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)

The letters in parentheses record the physical state of each substance. Use (s) for a solid, (l) for a liquid and (g) for a gas. Use (aq) whenever the substance is dissolved in water. That last label is short for aqueous, which simply means dissolved in water.

Two kinds of number show up in an equation, and they do completely different jobs. The large number in front of a formula is a coefficient, and it counts whole units of that substance. The small lowered number inside a formula is a subscript, and it counts atoms inside a single unit. You may change a coefficient whenever balancing requires it. You may never change a subscript, because a different subscript names a different substance. Carbon dioxide, CO₂, behaves nothing like carbon monoxide, CO.

Why the two sides must match

Atoms are never created or destroyed during a chemical change, and that statement is the law of conservation of matter. Bonds break and new bonds form, but the very same atoms turn up again in the products. So every element must appear in equal numbers on both sides of the arrow. An equation that passes that test is balanced.

The reliable way to get there is an atom inventory. List every element down the page, count it on the left, then count it on the right. Change one coefficient and count again. Two habits make the process a great deal faster. Begin with an element that appears in only one formula on each side. Save the element that appears in the most formulas for last, which is usually oxygen. Finish by reducing everything to the smallest whole numbers that still work.

Worked example · balance the burning of ethane

Ethane, C₂H₆, burns in oxygen to give carbon dioxide and water. The skeleton is C₂H₆ + O₂ → CO₂ + H₂O.

Count what one ethane unit holds

One C₂H₆ unit holds 2 C and 6 H. Set those two elements first.

Match carbon, then hydrogen

Two carbon atoms need 2 CO₂. Six hydrogen atoms need 3 H₂O.

Total the oxygen on the right

2 CO₂ holds 4 O. 3 H₂O holds 3 more. That is 7 O atoms, which takes 7/2 units of O₂.

Clear the fraction

Half units are not allowed, so double every coefficient.

2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(g) Check all three elements

Carbon: 4 on the left and 4 on the right. Hydrogen: 12 and 12. Oxygen: 14 on the left, and 8 + 6 = 14 on the right. It balances.

Figure 4.1 Methane combustion as an atom count. The coefficients 1, 2, 1 and 2 are the only set of small whole numbers that makes every row of the inventory match.

A note on fractions. A coefficient of 7/2 is not wrong in thermochemistry, where chemists often balance to exactly one mole of fuel. For ordinary stoichiometry, clear the fraction and keep whole numbers.

Key points

Coefficients are yours to change. Subscripts are not.
Balanced means every element has the same count on both sides.
The labels (s), (l), (g) and (aq) record the state of each substance.
Start with an element in only one formula per side. Leave oxygen for last.
Report the smallest whole numbers that work.

Interactive Balance equations by atom counting with live feedback. Open full ↑

Section 2Classifying reactions

One reaction, three equations

A huge share of chemistry happens in water. A reaction in water can be written three ways, and each way shows something the others hide. The molecular equation shows every compound as a whole formula. The complete ionic equation splits each dissolved salt, strong acid and strong base into its free ions. The net ionic equation keeps only the parts that change. Ions that start out free and end up free took no part at all, and those are the spectator ions you cross off.

Precipitation: a solid falls out

Pour two clear solutions together and a cloud of solid can suddenly appear. That solid is a precipitate, and the change is a precipitation reaction. To predict one in advance you need the solubility rules, which are short statements about which ionic solids dissolve well in water and which hardly dissolve at all.

Ion in the compound	Usually dissolves	Watch out for
NO₃^-, CH₃COO^-	Yes, every one	No common exceptions
Group 1 metals, NH₄⁺	Yes, every one	No common exceptions
Cl^-, Br^-, I^-	Yes, most of them	Ag⁺, Pb²⁺ and Hg₂²⁺ salts stay solid
SO₄^2-	Yes, most of them	Ba²⁺, Sr²⁺ and Pb²⁺ stay solid; Ca²⁺ barely dissolves
OH^-	Only group 1 and Ba²⁺	Ca²⁺ and Sr²⁺ barely dissolve; the rest are solid
CO₃^2-, PO₄^3-, S^2-	Only group 1 and NH₄⁺	Nearly all others stay solid

Using the table takes two simple moves. Swap the partners in the two compounds to see the possible new pairings, then test each new pair against the rules. Any pair the table calls insoluble is your precipitate.

Worked example · lead nitrate meets potassium iodide

Swap the partners

Pb²⁺ can pair with I^-, and K⁺ can pair with NO₃^-. Balance the charges to get PbI₂ and KNO₃.

Test both pairs

Every nitrate dissolves, so KNO₃ stays in solution. Iodides dissolve except with lead, so PbI₂ is the solid.

Write the molecular equation and balance it Pb(NO₃)₂(aq) + 2 KI(aq) → PbI₂(s) + 2 KNO₃(aq) Split everything dissolved into ions Pb²⁺(aq) + 2 NO₃^-(aq) + 2 K⁺(aq) + 2 I^-(aq) → PbI₂(s) + 2 K⁺(aq) + 2 NO₃^-(aq) Cross off the spectators

K⁺ and NO₃^- appear unchanged on both sides, so they drop out.

Pb²⁺(aq) + 2 I^-(aq) → PbI₂(s)

The net ionic equation is the whole story in six symbols. Charge balances too, since 2+ and two 1- ions add to zero.

Acids and bases trade a proton

An acid gives away a hydrogen ion, H⁺. A base accepts one. When an acid meets a base in water the two cancel out, and chemists call that an acid-base reaction, or neutralization. The products are water and a dissolved salt. A strong acid comes apart fully in water. The common ones are HCl, HBr, HI, HNO₃, H₂SO₄ and HClO₄. A weak acid only comes apart part way, like the acetic acid in vinegar.

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Strip out the Na⁺ and Cl^- spectators and a single short line remains. Every strong acid and strong base pairing boils down to exactly the same change.

H⁺(aq) + OH^-(aq) → H₂O(l)

Redox moves electrons

In the third family, electrons actually move from one atom to another. Chemists call that a redox reaction, which is short for oxidation and reduction. Burning fuel is redox, and so is iron rusting or a battery running your phone. Combustion is the rapid reaction of a fuel with oxygen. In a single displacement reaction, one element shoves another element out of its compound. Zinc metal does exactly that to the hydrogen in hydrochloric acid.

Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g)

Some reactions sit in two families at once, and the next section gives you a firm test for the redox family.

Figure 4.2 Run a new equation past all three questions. A reaction may answer yes to more than one, and burning a fuel in air is both a combustion and a redox change.

Key points

A net ionic equation drops every spectator ion and shows only the change.
Swap the partners, then check the solubility rules to find a precipitate.
Strong acid plus strong base always nets out to H⁺ + OH^- → H₂O.
Redox means electrons changed hands between atoms.
These families overlap, so one reaction can carry two labels.

Interactive Sort reactions into precipitation, acid-base and redox. Open full ↑

Section 3Oxidation numbers and redox

A bookkeeping charge

An oxidation number is the charge an atom would carry if every bond around it were completely ionic. You pretend each shared pair belongs entirely to the greedier atom, and then you count. The result is usually not the real charge on that atom at all. It is still worth the trouble, because any change in the number proves that electrons moved.

The rules, in order

Work down this list and stop as soon as a rule settles an atom, because the higher rules always win.

An atom in a free element is 0. In O₂, in Na metal and in S₈, every atom is 0.
A single atom ion takes its own charge. Na⁺ is +1 and S^2- is -2.
Fluorine in a compound is always -1.
Group 1 metals are +1. Group 2 metals are +2.
Hydrogen is +1 with a nonmetal and -1 with a metal, as in NaH.
Oxygen is -2, except in a peroxide like H₂O₂, where it is -1.
Other halogens are -1 unless bonded to oxygen or to fluorine.
The numbers sum to 0 in a neutral compound. In an ion they sum to the charge.

Rule 8 is the one that does the real work, because it lets you solve for the single atom you do not know yet.

Worked example · chromium in the dichromate ion

Find the oxidation number of chromium in Cr₂O₇^2-.

Pin down the atoms you can

Oxygen is -2 here, and there are seven of them. That is 7 × (-2) = -14.

Write the sum

Let x be the number for one chromium atom. Two of them plus the oxygen must equal the ion charge, so 2x + (-14) = -2.

Solve

2x = -2 + 14 = 12, so x = +6. Each chromium atom is +6.

Check on a second ion

In MnO₄^-, four oxygen atoms give -8. Then x - 8 = -1, so manganese is +7. In KMnO₄ the potassium is +1, so x + 1 - 8 = 0 and manganese is +7 again.

Which atom was oxidized?

Oxidation is a rise in oxidation number, which means the atom gave electrons away. Reduction is a drop in oxidation number, which means the atom took electrons in. Neither one can happen alone, since the electrons always have to go somewhere. The substance that accepts the electrons is the oxidizing agent, and the substance that hands them over is the reducing agent. Notice the twist in those names, because the oxidizing agent is itself the species that gets reduced.

You can split the whole change into two half-reactions, one for the loss and one for the gain. Zinc metal dropped into copper sulfate solution shows both halves clearly.

Zn(s) → Zn²⁺(aq) + 2 e^- and Cu²⁺(aq) + 2 e^- → Cu(s)

Worked example · smelting iron ore

Iron oxide reacts with carbon monoxide in a blast furnace.

Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g) Number every atom

In Fe₂O₃, oxygen is -2, so 2x - 6 = 0 and iron is +3. In CO, oxygen is -2, so carbon is +2. Free iron metal is 0. In CO₂, carbon is +4.

Follow each element across the arrow

Iron goes from +3 to 0, a drop of 3. Iron was reduced. Carbon goes from +2 to +4, a rise of 2. Carbon was oxidized.

Check that the electrons balance

Two iron atoms gain 3 electrons each, which is 6 electrons. Three carbon atoms lose 2 electrons each, which is also 6. The books close.

Name the agents

Fe₂O₃ took the electrons, so it is the oxidizing agent. CO gave them up, so it is the reducing agent.

Figure 4.3 One rise always pairs with one fall. The two electrons zinc releases are exactly the two a copper ion needs, so the total charge never changes.

Key points

Oxidation numbers are a count, not a measured charge.
Free elements are 0, and the numbers sum to the overall charge.
A rise is oxidation. A drop is reduction. They always come in pairs.
The oxidizing agent is the substance that gets reduced.
Electrons lost must equal electrons gained.

Interactive Assign oxidation numbers and spot what was oxidized and reduced. Open full ↑

Section 4Mole to mole stoichiometry

The coefficients are a ratio

A balanced equation is far more than a tidy record of a change. Its coefficients give you a ratio you can actually count with, so read them as moles. In the reaction that makes ammonia, one mole of N₂ reacts with three moles of H₂ and yields two moles of NH₃.

N₂(g) + 3 H₂(g) → 2 NH₃(g)

Any pair of those numbers is a mole ratio. From this single equation you can pull 1 mol N₂ per 3 mol H₂, or 2 mol NH₃ per 1 mol N₂, and several others. A mole ratio is simply a conversion factor, so write it as a fraction and let the units cancel. The study of these reacting amounts is called stoichiometry.

The road map

Here is the snag. A laboratory balance weighs in grams, but the equation speaks only in moles. So almost every amount problem walks the same four stop road. Start with grams of the substance you know, then divide by its molar mass to reach moles. Multiply by the mole ratio to cross over into moles of the substance you want. Finally multiply by that second molar mass to reach grams of your answer.

Notice that the mole ratio is the only step that ever crosses the arrow, since the two outer steps never leave their own substance. If a problem hands you moles instead of grams, you simply skip a step and begin further along the road.

Figure 4.4 The road map for any amount problem. Only the middle arrow uses the balanced equation, and the two end arrows are pure molar mass work.

Worked example · mass of ammonia from 25.0 g of nitrogen

Nitrogen reacts with plenty of hydrogen. What mass of NH₃ can form?

Start from the balanced equation

N₂ + 3 H₂ → 2 NH₃. The ratio you need is 2 mol NH₃ per 1 mol N₂.

Get the molar masses

N₂ is 2 × 14.007 = 28.014 g/mol. NH₃ is 14.007 + 3 × 1.008 = 17.031 g/mol.

Grams of A to moles of A

25.0 g ÷ 28.014 g/mol = 0.8924 mol N₂. The grams cancel and moles are left.

Moles of A to moles of B

0.8924 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 1.785 mol NH₃.

Moles of B to grams of B

1.785 mol × 17.031 g/mol = 30.4 g NH₃.

Check the significant figures

The given mass 25.0 g carries three significant figures, so the answer keeps three. The result is 30.4 g of NH₃.

Worked example · a pure mole to mole step

Propane burns by C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O. You have 1.50 mol of propane.

Oxygen needed

1.50 mol C₃H₈ × (5 mol O₂ / 1 mol C₃H₈) = 7.50 mol O₂.

Carbon dioxide formed

1.50 mol × (3 mol CO₂ / 1 mol C₃H₈) = 4.50 mol CO₂.

Water formed

1.50 mol × (4 mol H₂O / 1 mol C₃H₈) = 6.00 mol H₂O. No molar mass was needed at all.

Key points

Coefficients give mole ratios, not mass ratios.
Write every ratio as a fraction and cancel the units.
Grams to moles divides by molar mass. Moles to grams multiplies by it.
Only the middle step of the road map crosses the arrow.
The given value with the fewest significant figures sets the answer.

Interactive Follow the grams to moles to moles to grams road map. Open full ↑

Section 5Limiting reactant

When one ingredient runs out

Think about building sandwiches, where each one takes two slices of bread and one slice of cheese. You have 10 slices of bread and only 3 slices of cheese. The bread could cover five sandwiches, but the cheese stops you at three. Cheese is the ingredient that sets the limit, and four slices of bread are left on the counter.

Reactions behave in exactly the same way. The reactant that runs out first is the limiting reactant, and it alone decides how much product can form. Any reactant still sitting there afterward is present in excess, and that leftover material is the excess reactant.

How to find the limiter

Do not compare masses. Grams are a trap here, because a mole of one substance can weigh fifteen times as much as a mole of another. Compare what each reactant could make instead, by converting each starting amount into moles of the same product. The smallest answer is the real one, and the reactant that produced it is limiting.

Once you know the limiter, the rest is simply the road map from the previous section, using only the limiting amount. To find what remains, work out how much of the excess reactant the reaction actually consumed, then subtract that from what you started with.

Worked example · 10.0 g of hydrogen and 64.0 g of oxygen

The reaction is 2 H₂(g) + O₂(g) → 2 H₂O(g).

Molar masses

H₂ is 2.016 g/mol. O₂ is 32.00 g/mol. H₂O is 18.02 g/mol.

Moles of each reactant

10.0 g ÷ 2.016 g/mol = 4.96 mol H₂. 64.0 g ÷ 32.00 g/mol = 2.000 mol O₂.

Water each one could make

From hydrogen: 4.96 mol × (2 mol H₂O / 2 mol H₂) = 4.96 mol H₂O. From oxygen: 2.000 mol × (2 mol H₂O / 1 mol O₂) = 4.000 mol H₂O.

Pick the smaller one

Oxygen gives the smaller number, so oxygen is limiting. Only 4.000 mol of water can form, even though hydrogen could have supported more.

Mass of water

4.000 mol × 18.02 g/mol = 72.1 g H₂O.

Leftover hydrogen

Oxygen used 2.000 mol × (2 mol H₂ / 1 mol O₂) = 4.000 mol H₂. You began with 4.96 mol, so 4.96 - 4.00 = 0.96 mol is left. That is 0.96 mol × 2.016 g/mol = 1.9 g of H₂.

Check with conservation of mass

You started with 10.0 + 64.0 = 74.0 g. You end with 72.1 g of water plus 1.9 g of hydrogen, which is 74.0 g. Nothing went missing.

Look again at the masses in that example, because there was far more oxygen on the balance, yet oxygen still ran out first. The ratio inside the equation, and not the weight on the balance, is what really matters.

Figure 4.5 Each bar is the water that one reactant could make on its own. The lower bar sets a hard ceiling, so the shaded strip on the hydrogen bar can never be reached.

Why run an excess on purpose? In real labs the cheap reactant is often loaded in excess. Extra oxygen in a furnace pushes the fuel to burn completely, and the leftover gas simply leaves with the exhaust.

Key points

The limiting reactant runs out first and caps the product.
Never compare masses. Compare the product each reactant could make.
Convert both starting amounts to moles of the same product and take the smaller.
Leftover equals what you started with minus what the reaction used.
Total mass in must equal total mass out, leftovers included.

Interactive Run a reaction with set amounts and find what runs out first. Open full ↑

Section 6Theoretical, actual and percent yield

Three different yields

Stoichiometry tells you the most product you could possibly get, but real flasks rarely deliver it. Chemists keep three separate numbers apart so they can report how a run actually went.

Theoretical yield is the amount the limiting reactant allows, calculated from the road map, so it is a prediction. Actual yield is what you really collect, dry and weigh, so it is a measurement. Actual yield is almost always the smaller of the two. Some product clings to the glassware. Some reactant never finds a partner. A side reaction can steal material, or the reaction may reach equilibrium before it finishes. Percent yield puts the two numbers side by side.

percent yield = (actual yield ÷ theoretical yield) × 100%

Both yields must be expressed in the same unit before you divide. Grams over grams works, and moles over moles works, but you cannot mix the two.

Worked example · how the ammonia run really went

Section 4 predicted 30.4 g of NH₃ from 25.0 g of N₂. The flask gave 24.8 g.

Name each number

Theoretical yield is 30.4 g. Actual yield is 24.8 g.

Divide and scale

24.8 g ÷ 30.4 g = 0.8158. Multiply by 100% to get 81.6%.

Say what it means

You captured about 82 parts in every 100 that the equation promised. The other 18 parts stayed behind or never reacted.

Working the formula backward

The same relationship answers two other useful questions. If you know the usual percent yield, you can predict the real haul before you even begin. If you need a set amount of product, you can work out how much starting material to weigh.

Worked example · planning around a known yield

A synthesis step reliably runs at 72.0%. Stoichiometry says the batch should give 45.0 g.

Predict the actual yield

45.0 g × 0.720 = 32.4 g of product is what you should expect.

Turn the question around

Suppose you need 32.4 g in hand. Then the theoretical yield has to be 32.4 g ÷ 0.720 = 45.0 g, which tells you how much reactant to weigh out.

Chain two steps

A second step at 80.0% follows the first. The overall yield is 0.720 × 0.800 = 0.576, or 57.6%. Losses multiply, so long routes bleed material fast.

One warning before you go. A percent yield above 100% is not a triumph, because it means the measurement itself is wrong. The product is probably still damp with solvent, or some unreacted solid came along with it. Dry the sample thoroughly and then weigh it again.

Figure 4.6 Percent yield is one bar divided by the other. The dashed gap is product that the equation allowed but the run never delivered.

Atom economy. Percent yield grades the run. Atom economy grades the route. It asks what fraction of the mass of all the reactants ends up in the product you want. A route with high atom economy wastes less, even before you step into the lab.

Key points

Theoretical yield is predicted. Actual yield is weighed.
Percent yield is actual divided by theoretical, times 100%.
Both yields must share the same unit before you divide.
Multiply the yields of chained steps to get the overall yield.
Over 100% means a measurement problem, such as a damp product.

Interactive Compare what should form with what really came out. Open full ↑

The chapter in one breath

A balanced equation keeps every element equal on both sides, because atoms are never lost.
Change coefficients to balance. Never touch a subscript.
A net ionic equation drops the spectator ions and shows only the real change.
Solubility rules predict a precipitate. Acid plus base nets out to water. Redox shifts oxidation numbers.
An oxidation number rise is oxidation. A drop is reduction. The two always pair up.
Coefficients are mole ratios, so they convert moles of one substance into moles of another.
Every mass problem walks the same road: grams, moles, moles, grams.
The limiting reactant is the one that makes the least product, and it sets the theoretical yield.
Percent yield is actual divided by theoretical, times 100%, and it is almost never 100.

ReferenceWords worth knowing

These are the terms this chapter leans on. Come back to the list whenever an equation stops making sense.

acid-base reaction: A reaction in which an acid transfers H⁺ to a base, usually giving water and a dissolved salt. Also called neutralization.
actual yield: The mass or moles of product you really collect and weigh at the end of a run.
balanced equation: A chemical equation whose coefficients give every element the same atom count on both sides of the arrow.
coefficient: The whole number written in front of a formula. It counts units of that substance and may be changed to balance an equation.
combustion: The rapid reaction of a fuel with oxygen, releasing heat and light. Hydrocarbon combustion gives CO₂ and H₂O.
excess reactant: A reactant still present after the reaction stops, because the limiting reactant ran out first.
limiting reactant: The reactant that is used up first. It sets the maximum amount of product that can form.
mole ratio: A conversion factor built from two coefficients of a balanced equation, such as 2 mol NH₃ per 1 mol N₂.
net ionic equation: An equation in which spectator ions are removed, leaving only the species that change.
oxidation: An increase in oxidation number, which means the atom lost electrons.
oxidation number: The charge an atom would have if every bond in its compound were fully ionic. A bookkeeping device, not a measured charge.
oxidizing agent: The substance that accepts electrons in a redox reaction. It is the species that gets reduced.
percent yield: Actual yield divided by theoretical yield, multiplied by 100%. A measure of how well a run performed.
precipitate: An insoluble solid that forms and separates when two solutions are mixed.
reducing agent: The substance that donates electrons in a redox reaction. It is the species that gets oxidized.
reduction: A decrease in oxidation number, which means the atom gained electrons.
solubility rules: A short set of statements that predict whether a given ionic compound dissolves well in water.
spectator ion: An ion that is present in solution before and after the reaction and takes no part in the change.
stoichiometry: The use of a balanced equation to relate amounts of reactants and products, in moles, mass or volume.
theoretical yield: The maximum product the limiting reactant allows, calculated from the balanced equation.

Check yourself

1. Why may you change a coefficient while balancing, but never a subscript?

A coefficient counts units; a subscript defines the substance. Changing 2 CO₂ to 3 CO₂ just takes more of the same gas. Changing CO₂ to CO₃ invents a different substance that was never in the reaction.

2. Balance this equation: Fe + O₂ → Fe₂O₃.

4 Fe + 3 O₂ → 2 Fe₂O₃ One Fe₂O₃ needs 3 O, so you need 3/2 O₂. Double everything to clear the fraction: Fe is 4 = 2 × 2, and O is 3 × 2 = 6 = 2 × 3.

3. Write the net ionic equation for AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq), and name the spectators.

Ag⁺(aq) + Cl^-(aq) → AgCl(s); Na⁺ and NO₃^- are spectators. NaNO₃ stays fully dissolved, so its two ions appear unchanged on both sides and cancel out.

4. Mix Ba(NO₃)₂(aq) with Na₂SO₄(aq). Does a solid form?

Yes. BaSO₄ precipitates. Most sulfates dissolve, but barium sulfate is one of the exceptions. The other new pair, NaNO₃, is soluble, so the net ionic equation is Ba²⁺ + SO₄^2- → BaSO₄(s).

5. What is the oxidation number of sulfur in SO₄^2-?

+6 Each of the four oxygen atoms is -2, giving -8. The sum must equal the ion charge, so x + (-8) = -2. Then x = -2 + 8 = +6.

6. What is the oxidation number of nitrogen in NH₄⁺?

-3 Hydrogen bonded to a nonmetal is +1, and four of them give +4. The sum must equal +1, so x + 4 = +1 and x = -3.

7. In Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g), what is oxidized and what is the oxidizing agent?

Zinc is oxidized; HCl is the oxidizing agent. Zn goes from 0 to +2, a rise of 2, so it lost electrons. Hydrogen drops from +1 to 0, so it was reduced, which makes HCl the oxidizing agent. Chlorine stays at -1 throughout.

8. How many moles of O₂ are needed to burn 2.40 mol of CH₄?

4.80 mol O₂ The balanced equation is CH₄ + 2 O₂ → CO₂ + 2 H₂O. So 2.40 mol CH₄ × (2 mol O₂ / 1 mol CH₄) = 4.80 mol O₂.

9. What mass of CO₂ forms when 15.0 g of propane, C₃H₈, burns completely?

44.9 g CO₂ C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O. Molar mass of C₃H₈ is 44.10 g/mol, so 15.0 ÷ 44.10 = 0.3401 mol. Then 0.3401 × 3 = 1.020 mol CO₂, and 1.020 × 44.01 g/mol = 44.9 g.

10. You mix 3.00 mol Al with 4.00 mol Cl₂ by 2 Al + 3 Cl₂ → 2 AlCl₃. Which reactant limits, and how much AlCl₃ forms?

Cl₂ limits; 2.67 mol AlCl₃ forms. Aluminum could make 3.00 × (2/2) = 3.00 mol AlCl₃. Chlorine could make 4.00 × (2/3) = 2.67 mol. The smaller value wins. Aluminum used is 4.00 × (2/3) = 2.67 mol, so 0.33 mol Al is left over.

11. A run has a theoretical yield of 8.50 g and you collect 7.23 g. What is the percent yield?

85.1% Divide actual by theoretical: 7.23 g ÷ 8.50 g = 0.8506. Multiply by 100% to get 85.1%, keeping three significant figures.

12. Your calculation gives a percent yield of 104%. What should you suspect?

The product is impure or still wet, so the mass is too high. Nothing can beat the amount the limiting reactant allows. Solvent left in the solid, or unreacted starting material mixed in, both add mass. Dry the sample and weigh it again.

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