Chemistry · Ch 2 — Introduction to Analytical Chemistry

Percent composition and empirical formula

2.4.1

Percent composition and empirical formula

Compounds form when different elements combine chemically in fixed proportions, and quantitatively determining how much of each constituent element is present, by suitable analytical methods, gives the percent elemental composition of the compound. If the measured percentages of the known elements do not add up to 100%, the shortfall is conventionally attributed to oxygen (on the reasoning that oxygen is easy to lose or hard to measure directly in many organic analyses). From the percent composition, the relative number of moles — and hence the simplest whole-number ratio of atoms — of each constituent element can be calculated; this simplest whole-number ratio is called the empirical formula of the compound. The true molecular formula can then be obtained from the empirical formula, provided the compound's molar mass has also been measured by some convenient experimental method, because the molecular formula is always a whole-number multiple of the empirical formula. The standard six-step procedure — checking the percentages sum to 100%, converting percent to grams, converting grams to moles using atomic masses, dividing every mole value by the smallest one to get a simple ratio, writing the empirical formula from that ratio, and finally scaling it to the molecular formula using the ratio of mola …

Misc Problem 2.9Finding the empirical and molecular formula of a compound from its percent composition (H, C, Cl)

Worked out. Worked example: a compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine by mass, and has a molar mass of 98.96 g. Atomic masses used: H = 1.008, C = 12.00 (12.01 used in the division step), Cl = 35.453. Step I — check the percentages sum to 100%: $4.07 + 24.27 + 71.65 = 99.99 \approx 100$ , so no oxygen needs to be assumed. Step II — treat 100 g of compound as the sample, so it contains 4.07 g H, 24.27 g C and 71.65 g Cl. Step III — convert each mass to moles by dividing by its atomic mass: moles H $= 4.07/1.008 = 4.04$ ; moles C $= 24.27/12.01 = 2.0225$ ; moles Cl $= 71.65/35.453 = 2.021$ . Step IV — divide every mole value by the smallest one (2.021, chlorine's): this gives a whole-number ratio of H : C : Cl $= 2 : 1 : 1$ (if a ratio is not already close to whole numbers, every value is multiplied by the same small integer until it is). Step V — write the empirical formula using these ratios as subscripts: CH2Cl. Step VI — find the molecular formula: (a) empirical formula mass of CH2Cl $= 12.01 + 2(1.008) + 35.453 = 49.48$ g; (b) divide the given molar mass by this empirical formula mass, $98.96/49.48 = 2$ , so the multiplying factor $r = 2$ ; (c) multiply every su …

Misc Problem 2.10Finding the molecular formula of a copper–sulfur–oxygen compound

Worked out. Worked example: a compound with molar mass 159 is found to contain 39.62% copper and 20.13% sulfur (atomic masses Cu = 63, S = 32, O = 16). Percent Cu + percent S $= 39.62 + 20.13 = 59.75\%$ , which is less than 100%, so the remaining $100 - 59.75 = 40.25\%$ is taken as oxygen. Moles: Cu $= 39.62/63 = 0.629$ ; S $= 20.13/32 = 0.629$ ; O $= 40.25/16 = 2.516$ . Dividing every value by the smallest (0.629) gives the ratio Cu : S : O $= 1 : 1 : 4$ , so the empirical formula is CuSO4. Its formula mass $= 63 + 32 + 16 \times 4 = 159$ , which equals the given molar mass, so the molecular formula is the same as th …

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