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Chemistry · Ch 1 — Some Basic Concepts of Chemistry

Percentage Composition

1.9

Percentage Composition

So far we were dealing with the number of entities present in a given sample; many a time, information regarding the percentage of a particular element present in a compound is required. Suppose an unknown or new compound is given to you — the first question you would ask is: what is its formula, and in what ratio are its constituents present? For known compounds too, such information provides a check on purity: one can check the purity of a given sample by analysing this data.

Figure 1.11One mole of various substances -- water, sodium chloride, glucose, and carbon -- each containing 6.022x10^23 elementary entities, with their respective molar masses.
Fig. 1.11 — One mole of various substances -- water, sodium chloride, glucose, and carbon -- each containing 6.022x10^23 elementary entities, with their respective molar masses.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Fig. 1.11 shows four everyday substances side by side, each in the amount that makes up exactly one mole: 18.02 g of water, 58.5 g of sodium chloride, 180.16 g of glucose, and 12.01 g of carbon. Even though these masses are all different, each sample contains the same number of elementary entities -- 6.022x10^23, Avogadro's number -- which is the whole point o …

The mass per cent of each element is worked out from the compound's formula and molar mass:

Mass % of an element=mass of that element in the compound×100molar mass of the compound\text{Mass \% of an element} = \frac{\text{mass of that element in the compound} \times 100}{\text{molar mass of the compound}}

Worked case 1: water (H₂O)

Molar mass of water = 18.02 g.

Mass % of hydrogen=2×1.00818.02×100=11.18\text{Mass \% of hydrogen} = \frac{2 \times 1.008}{18.02} \times 100 = 11.18

Mass % of oxygen=16.0018.02×100=88.79\text{Mass \% of oxygen} = \frac{16.00}{18.02} \times 100 = 88.79

Worked case 2: ethanol (C₂H₅OH)

Molar mass of ethanol = (2 × 12.01) + (6 × 1.008) + 16.00 = 46.068 g.

Mass per cent of carbon=24.02 g46.068 g×100=52.14%\text{Mass per cent of carbon} = \frac{24.02\ \text{g}}{46.068\ \text{g}} \times 100 = 52.14\%

Mass per cent of hydrogen=6.048 g46.068 g×100=13.13%\text{Mass per cent of hydrogen} = \frac{6.048\ \text{g}}{46.068\ \text{g}} \times 100 = 13.13\%

Mass per cent of oxygen=16.00 g46.068 g×100=34.73%\text{Mass per cent of oxygen} = \frac{16.00\ \text{g}}{46.068\ \text{g}} \times 100 = 34.73\%

Tip

Always check that the percentages add up to approximately 100%. For ethanol: 52.14 + 13.13 + 34.73 = 100.00%. …

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