What is the percent composition of $H_{2} S$ $H_2S$ ?

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What is the percent composition of $H_{2} S$ $H_2S$ ?

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Answer 1 · 2016-04-02T15:23:31

$5.92 % H$ $5.92% H$ , $94.08 % S$ $94.08% S$

Explanation:

Total mass of a compound is found by adding up the mass of all the atoms. Using mass numbers from the RSC periodic table, the total mass of the compound $H_{2} S$ $H_2S$ is

$2 \cdot 1.008 g m o l^{- 1} + 1 \cdot 32.06 g m o l^{- 1} = 34.076 g m o l^{- 1}$ $2 * 1.008gmol^-1 + 1 * 32.06gmol^-1 = 34.076gmol^-1$

Percentage mass of an element in a compound is given by

$\frac{M_{e}}{M_{t}} \cdot 100 %$ $M_e/M_t * 100%$

where $M_{e}$ $M_e$ is the mass of all the atoms of the element and $M_{t}$ $M_t$ is the total mass of the compound.

Percentage mass of hydrogen therefore is

$\frac{2.016 g m o l^{- 1}}{34.076 g m o l^{- 1}} \cdot 100 % = 5.92 %$ $(2.016gmol^-1)/(34.076gmol^-1) * 100% = 5.92%$

And, likewise, percentage mass of sulfur is

$\frac{32.06 g m o l^{- 1}}{34.076 g m o l^{- 1}} \cdot 100 % = 94.08 %$ $(32.06gmol^-1)/(34.076gmol^-1) * 100% = 94.08%$

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What is the percent composition of $H_{2} S$ $H_2S$ ?

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