Elemental Composition of Pure Substances
Introduced| English | Meaning |
|---|---|
| percent composition/pəˈsent ˌkɒmpəˈzɪʃn/ | The mass of each element in a compound expressed as a percentage of the compound's total mass. |
| empirical formula/emˈpɪrɪkl ˈfɔːmjʊlə/ | A formula showing the simplest whole-number ratio of atoms of each element in a compound. |
| molecular formula/məˈlekjʊlə ˈfɔːmjʊlə/ | A formula showing the actual numbers of atoms of each element in one molecule. |
What is it made of?
- Break any compound down and you find fixed proportions by mass.
- Water is always 11% hydrogen and 89% oxygen by weight.
- Those percentages never change, no matter the sample size.
- From the percentages you can rebuild the formula.
Percent by mass
- The percent composition 百分组成 is each element's share of the total mass:
- For a formula, work it out from one mole and the molar masses.
Finding the empirical formula 实验式: mass to moles, divide by the smallest, read the ratio
In water ($\text{H}_2\text{O}$, molar mass 18), what percent of the mass is oxygen (16)? (to nearest whole %)
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$16/18 \times 100 \approx 89\%$.
The empirical formula
- The empirical formula is the simplest whole-number ratio of atoms.
- Convert each element's mass to moles, then divide by the smallest.
- Glucose has the empirical formula $\text{CH}_2\text{O}$.
The empirical formula of a compound gives...
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Empirical = simplest ratio; the molecular formula gives the actual counts.
From empirical to molecular
- The molecular formula 分子式 is a whole-number multiple of the empirical one.
- Divide the true molar mass by the empirical mass to get that multiple $n$.
- Glucose: empirical $\text{CH}_2\text{O}$ ($30$), molecular $\text{C}_6\text{H}_{12}\text{O}_6$ ($180$), so $n = 6$.
A compound has empirical mass 30 and molar mass 90. The multiple $n$ for its molecular formula?
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$n = 90/30 = 3$.
The molecular formula is always a whole-number multiple of the empirical formula.
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By definition, molecular = empirical × n for some whole number n.
A compound is 40% C, 6.7% H, and 53.3% O by mass. Find its empirical formula.
- Moles per 100 g: C $= 40/12 = 3.33$, H $= 6.7/1 = 6.7$, O $= 53.3/16 = 3.33$.
- Divide by the smallest ($3.33$): C 1, H 2, O 1, giving $\text{CH}_2\text{O}$.
Percent composition to formula
Turn a compound's percent composition into its empirical formula, step by step.
Order the steps to find an empirical formula from mass percents.
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Grams → moles → divide by smallest → whole-number ratio.
After dividing by the smallest, you get a ratio of 1:1.5. You should...
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A .5 is real -- scale up to clear it, giving whole numbers 2:3.
Percent composition gives the empirical formula, not always the molecular one — you need the molar mass to find the multiple. When dividing moles by the smallest, round only near-whole results ($3.98 \to 4$); a stubborn $.5$ means multiply everything (by 2), never round it away. Start from a $100\ \text{g}$ sample so percentages become grams directly.
Percent composition is each element's fraction of the total mass. Turn those masses into moles and divide by the smallest to get the empirical formula (simplest ratio). The molecular formula is that empirical unit times $n$, where $n$ is the true molar mass divided by the empirical mass.