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a sample of a compound containing hydrogen, carbon, and sulfur is reacted with oxygen. the carbon content in the carbon dioxide obtained is 41.34% and the hydrogen content in the water obtained is 3.47%. determine the empirical formula of the compound.

Answer :

The empirical formula of the compound is C₂H₂S if the carbon content in the carbon dioxide obtained is 41.34% and the hydrogen content in the water obtained is 3.47%.

If the percentage of carbon content is 41.34% and hydrogen content is 3.47%, we can determine the percentage of sulfur content as follows;

percentage of sulfur = 100% - 41.34% - 3.47% = 55.19%

Now considering 100g of the compound;

Carbon = (41.34/100) × 100 = 41.34 g

Hydrogen = (3.47/100) × 100 = 3.47 g

Sulfur = (55.19/100) × 100 = 55.19 g

Now we can determine the moles of carbon, hydrogen, and sulfur by using their molar mass as follows;

Carbon: 41.34 ÷ 12.011 = 3.44 moles (Moles = mass ÷ molar mass)

Hydrogen: 3.47 ÷ 1.00784 = 3.44 moles

Sulfur: 55.19 ÷ 32.065 = 1.72 moles

Now by dividing each of the mole values by the smallest determined number of moles (1.72 moles), we can calculate the empirical formula as follows;

Carbon = 3.44 / 1.72 = 2

Hydrogen = 3.44 / 1.72 = 2

Sulfur = 1.72 / 1.72 = 1

Therefore the empirical formula of the compound will be C₂H₂S

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