How do you calculate the percent composition of water?

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How do you calculate the percent composition of water?

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Answer 1 · 2016-04-27T16:00:40

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Explanation:

For water we have to calculate it % of Oxygen and % of Hydrogen

Its long calculation to do it here but this video will help

Answer 2 · 2016-04-27T16:47:29

Since water is $H_{2} O$ $"H"_2"O"$ (molar mass is $18.015 g/mol$ $"18.015 g/mol"$ ), it implies we have a percent composition of $H$ $"H"$ and one for $O$ $"O"$ .

Given that chemical formula, we have two equivalents of $H$ $"H"$ , for which we can use its molar mass, $M_{H} = 1.0079 g/mol$ $M_"H" = "1.0079 g/mol"$ . We also have one equivalent of $O$ $"O"$ , for which we can use its molar mass, $M_{O} = 15.999 g/mol$ $M_"O" = "15.999 g/mol"$ .

The percent composition (by mass) of $H$ $"H"$ is written in terms of a unified ratio of the molar masses is

$% H = \frac{2 \times M_{H}}{M_{H_{2} O}} \times 100 %$ $%"H" = (2xxM_"H")/(M_("H"_2"O"))xx100%$

$= \frac{2 \times 1.0079 g/mol}{18.015 g/mol} \times 100 %$ $= (2xx"1.0079 g/mol")/("18.015 g/mol")xx100%$

$\approx 11.19 % H$ $~~ color(blue)(11.19% "H")$

Similarly for $O$ $"O"$ , we have

$% O = \frac{1 \times M_{O}}{M_{H_{2} O}} \times 100 %$ $%"O" = (1xxM_"O")/(M_("H"_2"O"))xx100%$

$= \frac{15.999 g/mol}{18.015 g/mol} \times 100 %$ $= ("15.999 g/mol")/("18.015 g/mol")xx100%$

$\approx 88.81 % O$ $~~ color(blue)(88.81% "O")$

Thus, the percent composition of water, by mass, is $11.19 %$ $11.19%$ hydrogen and $88.81 %$ $88.81%$ oxygen.

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How do you calculate the percent composition of water?

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