Chemical Calculations: Ratios, Empirical Formulae & Moles

Why chemical calculations use ratios

Chemistry links quantities that look different: the mass measured on a balance, the amount measured in moles and the numbers of particles shown by a chemical equation. Ratios connect them.

A chemical formula gives a ratio within a substance. For example, $\mathrm{H_2O}$ contains hydrogen and oxygen atoms in a $2:1$ ratio. A balanced equation gives ratios between substances. In

$2\mathrm{H_2}+\mathrm{O_2}\rightarrow2\mathrm{H_2O}$

the coefficients give the mole ratio $2:1:2$. This means 2 mol of hydrogen reacts with 1 mol of oxygen to form 2 mol of water.

Balancing equations by counting atoms

A chemical reaction rearranges atoms; it does not create or destroy them. A symbol equation must therefore contain the same number of atoms of every element on both sides.

Start with correct chemical formulae, count each type of atom on both sides, and place whole-number coefficients in front of formulae until the counts match. For example:

$\mathrm{Mg}+\mathrm{O_2}\rightarrow\mathrm{MgO}$

There are two oxygen atoms on the left but only one on the right. Placing 2 before $\mathrm{MgO}$ gives two oxygen atoms on the right, but now two magnesium atoms are needed on the left:

$2\mathrm{Mg}+\mathrm{O_2}\rightarrow2\mathrm{MgO}$

The balanced coefficients should normally be the smallest possible whole numbers. Never balance an equation by altering a subscript: changing $\mathrm{MgO}$ to $\mathrm{MgO_2}$ would describe a different substance rather than a different amount of magnesium oxide.

Converting between mass and moles

A mole is a chemical counting unit. The mass of 1 mol of a substance in grams is numerically equal to its relative particle mass. For an element this comes from its relative atomic mass, $A_r$; for a compound it comes from its relative formula mass, $M_r$.

Add the $A_r$ values for every atom shown in the formula. Using $A_r(\mathrm{Mg})=24$ and $A_r(\mathrm{O})=16$:

$M_r(\mathrm{MgO})=24+16=40$

The molar mass of magnesium oxide is therefore $40\ \mathrm{g\,mol^{-1}}$. The conversion is

$n=\frac{m}{M}$

where $n$ is amount in mol, $m$ is mass in g and $M$ is molar mass in $\mathrm{g\,mol^{-1}}$. Rearranging gives $m=nM$.

For example, $10.0\ \mathrm{g}$ of magnesium oxide is

$n=\frac{10.0}{40}=0.250\ \mathrm{mol}.$

This conversion matters because formulae and equations describe particle or mole ratios, not direct mass ratios.

Finding an empirical formula from masses

An empirical formula is the simplest whole-number ratio of atoms. Experimental masses must first be converted to moles because different atoms have different masses.

Use this sequence:

1. Write the mass of each element.
2. Calculate moles using $n=m/A_r$.
3. Divide every mole value by the smallest value.
4. If necessary, multiply all the ratios by the same small whole number.
5. Use the resulting whole numbers as subscripts in the empirical formula.

Suppose a compound contains $4.8\ \mathrm{g}$ of magnesium and $3.2\ \mathrm{g}$ of oxygen, with $A_r(\mathrm{Mg})=24$ and $A_r(\mathrm{O})=16$:

$n(\mathrm{Mg})=\frac{4.8}{24}=0.20\ \mathrm{mol}$

$n(\mathrm{O})=\frac{3.2}{16}=0.20\ \mathrm{mol}$

Dividing both by $0.20$ gives $1:1$, so the empirical formula is $\mathrm{MgO}$.

A ratio does not always immediately appear as whole numbers. For instance, $1:1.5$ becomes $2:3$ when both values are multiplied by 2. Multiplying must be applied to every value in the ratio.

Using percentage composition

Percentage composition can be handled as a mass calculation by imagining a $100\ \mathrm{g}$ sample. A compound that is 40% sulfur and 60% oxygen can be treated as containing $40\ \mathrm{g}$ of sulfur and $60\ \mathrm{g}$ of oxygen in a $100\ \mathrm{g}$ sample. Those masses can then be divided by the relevant $A_r$ values and simplified in exactly the same way as experimental masses.

More generally, the percentage by mass of an element is

$\text{percentage by mass}=\frac{\text{mass contribution of the element in the formula}}{M_r\text{ of the compound}}\times100.$

The mass contribution must include the number of atoms of that element. In $\mathrm{CO_2}$, for example, oxygen contributes $2\times16$, not just 16.

Multistep reacting-mass calculations

A typical quantitative calculation moves through three stages:

$\text{mass of known substance}\rightarrow\text{moles of known substance}\rightarrow\text{moles of required substance}\rightarrow\text{required mass}.$

For the equation

$2\mathrm{Mg}+\mathrm{O_2}\rightarrow2\mathrm{MgO},$

suppose $12\ \mathrm{g}$ of magnesium reacts completely. Using molar masses $24\ \mathrm{g\,mol^{-1}}$ for Mg and $40\ \mathrm{g\,mol^{-1}}$ for MgO:

• moles of Mg $=12/24=0.50\ \mathrm{mol}$;
• the equation gives an Mg:MgO mole ratio of $2:2$, or $1:1$, so $0.50\ \mathrm{mol}$ of MgO forms;
• mass of MgO $=0.50\times40=20\ \mathrm{g}$.

Notice that the coefficient ratio was used only after converting mass to moles. The numerical masses do not follow the coefficients because magnesium and magnesium oxide have different molar masses.

Percentages in multistep calculations

Many chemical percentages have the same mathematical structure:

$\text{percentage}=\frac{\text{part}}{\text{whole}}\times100.$

For percentage yield, the part is the actual amount of product obtained and the whole is the maximum theoretical amount:

$\text{percentage yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times100.$

The theoretical yield may itself require a mass-to-moles conversion, a mole ratio and a conversion back to mass. Keep unrounded calculator values during these stages and round only the final answer appropriately.

Keeping calculations reliable

Write the formula or balanced equation first, show each conversion and attach units to measured quantities. A useful dimensional check is

$\frac{\mathrm{g}}{\mathrm{g\,mol^{-1}}}=\mathrm{mol}.$

Finally, ask whether the result is chemically and numerically sensible: an empirical formula needs whole-number subscripts, a balanced equation must conserve every element, and an ordinary percentage should not exceed 100% unless the context explicitly permits it.

Core conversions

$n=\frac{m}{M}\qquad m=nM$

Use mass in g and molar mass in $\mathrm{g\,mol^{-1}}$. Calculate $M_r$ by adding the $A_r$ values for every atom in the formula.

Balance equations

• Use correct formulae first.
• Count each element on both sides.
• Alter coefficients only.
• Finish with the smallest whole-number ratio.

Coefficients give mole ratios, not mass ratios.

Empirical formula from mass or percentage

1. If given percentages, assume a $100\ \mathrm{g}$ sample.
2. For each element, calculate $\text{mass}/A_r$.
3. Divide all answers by the smallest.
4. Multiply every ratio if needed to obtain whole numbers.
5. Write these numbers as subscripts.

Reacting-mass route

$\text{mass}\rightarrow\text{moles}\xrightarrow{\text{coefficient ratio}}\text{moles}\rightarrow\text{mass}$

Balance the equation before using its ratio.

Percentages

$\text{percentage}=\frac{\text{part}}{\text{whole}}\times100$

$\text{percentage yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times100$

For percentage composition, include every atom of the chosen element in its mass contribution. Retain full calculator values during multistep work and round at the end.