Law Applications

Choose the Law from Data

The type of data in a problem indicates which basic law of chemistry applies. The problem may involve total mass, the composition of one compound, two compounds made from the same elements, or gas volumes under the same conditions.

Stoichiometric calculations use factors from balanced equations to relate amounts of substances. Before calculating, we need to check the balanced equation, the substance or quantity to calculate, and the units of the given data.

The following list connects each type of data with the appropriate law.

If every substance in the system is counted, use conservation of mass.
If one pure compound is being analyzed, use constant composition.
If two different compounds are made from the same two elements, use multiple proportions.
If all compared substances are gases at the same temperature and pressure, use combining volumes.

Mass and Volume Patterns

Before calculating, identify which quantity stays fixed and which quantities are being compared. That choice determines the law and ratio used in the calculation. The patterns below separate mass ratios from volume ratios, because the two follow different laws.

Reaction System

Suppose $10.00\ \mathrm{g}$ of zinc reacts with $2.00\ \mathrm{g}$ of sulfur. After the reaction, $6.08\ \mathrm{g}$ of zinc sulfide forms and $5.92\ \mathrm{g}$ of zinc remains.

\mathrm{Zn(s)} + \mathrm{S(s)} \rightarrow \mathrm{ZnS(s)}

The reacting zinc is not $10.00\ \mathrm{g}$ , because some zinc is left over.

\begin{aligned} &m_{\mathrm{Zn\ reacted}} = 10.00 - 5.92 = 4.08\ \mathrm{g} \\ &m_{\mathrm{reactants\ reacted}} = 4.08 + 2.00 \\ &= 6.08\ \mathrm{g} \end{aligned}

The mass of the reactants that actually reacted equals the product mass. If the whole container is counted, the mass before and after is also $12.00\ \mathrm{g}$ .

One Compound Composition

Calcium oxide, $\mathrm{CaO}$ , can be checked with two different samples. The first sample has mass $2.8\ \mathrm{g}$ and contains $0.8\ \mathrm{g}$ of oxygen. The second sample has mass $3.5\ \mathrm{g}$ and contains $1.0\ \mathrm{g}$ of oxygen.

\begin{aligned} m_{\mathrm{Ca,1}} &= 2.8 - 0.8 = 2.0\ \mathrm{g} \\ m_{\mathrm{Ca,2}} &= 3.5 - 1.0 = 2.5\ \mathrm{g} \\ 2.0:0.8 &= 2.5:1 \\ 2.5:1.0 &= 2.5:1 \end{aligned}

The sample masses may differ, but the calcium-to-oxygen mass ratio stays the same. That is the pattern of constant composition.

Comparing Different Compounds Made from the Same Elements

Water and hydrogen peroxide are both made from hydrogen and oxygen, but their formulas are different.

\mathrm{H_2O}\quad\text{and}\quad\mathrm{H_2O_2}

With $A_r\mathrm{(H)} = 1$ and $A_r\mathrm{(O)} = 16$ , the hydrogen mass in both compounds can be made the same: $2\ \mathrm{g}$ .

\begin{aligned} \mathrm{H_2O} &: 2\ \mathrm{g\ H} + 16\ \mathrm{g\ O} \\ \mathrm{H_2O_2} &: 2\ \mathrm{g\ H} + 32\ \mathrm{g\ O} \\ m_{\mathrm{O}} &: 16:32 = 1:2 \end{aligned}

Because the hydrogen mass is already the same, compare the oxygen mass. The ratio $1:2$ is a simple whole-number ratio, matching the law of multiple proportions.

Gases Under Matching Conditions

For example, ammonia decomposition:

2\mathrm{NH_3(g)} \rightarrow 3\mathrm{H_2(g)} + \mathrm{N_2(g)}

If $600\ \mathrm{mL}$ of ammonia decomposes, then $2$ parts equal $600\ \mathrm{mL}$ and $1$ part equals $300\ \mathrm{mL}$ .

\begin{aligned} V_{\mathrm{H_2}} &= 3 \times 300 = 900\ \mathrm{mL} \\ V_{\mathrm{N_2}} &= 1 \times 300 = 300\ \mathrm{mL} \end{aligned}

A balanced equation has the same number of atoms on the reactant and product sides. At the same temperature and pressure, gas volume ratios follow the equation coefficients.

When One Problem Uses Two Laws

One problem can use more than one law. Product mass and leftover reactant use conservation of mass, but if the problem then asks for element composition in the same compound, that part uses constant composition. Do not force one law to answer every question.

Define the system boundary, then choose the part of the data that is actually being compared.

If the problem asks about total mass, define the system boundary first.
If it asks about one compound's composition, compare the masses of its elements.
If it compares two compounds from the same elements, hold one element fixed before comparing the other.
If it asks about gas volume, confirm that all gases are measured at the same temperature and pressure.

Methane Combustion

Methane burns according to this balanced equation.

\begin{aligned} \mathrm{CH_4(g)} + 2\mathrm{O_2(g)} &\rightarrow \mathrm{CO_2(g)} \\ &\quad + 2\mathrm{H_2O(g)} \end{aligned}

If $50\ \mathrm{mL}$ of methane gas reacts completely and every gas is measured at the same temperature and pressure, what volume of $\mathrm{O_2}$ is needed and what volume of $\mathrm{CO_2}$ forms?

The coefficients give the volume ratio $\mathrm{CH_4:O_2:CO_2:H_2O} = 1:2:1:2$ . Therefore:

\begin{aligned} V_{\mathrm{O_2}} &= 2 \times 50 = 100\ \mathrm{mL} \\ V_{\mathrm{CO_2}} &= 1 \times 50 = 50\ \mathrm{mL} \end{aligned}

The answer is $100\ \mathrm{mL}$ of oxygen gas and $50\ \mathrm{mL}$ of carbon dioxide gas. If the water product has condensed into liquid, its volume is not read directly from the gas ratio.

The laws of definite and multiple proportions show mass patterns that support the idea that atoms combine in fixed ratios.