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12. A 2.65 g compound is made of 0.70 g of chromium, 0.65 g of sulfur and the rest is oxygen. The molar mass is 392.21 g. Find the molecular formula of the compound.

User Jayabal
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Answer:

To find the molecular formula, we need to determine the empirical formula and then use the molar mass to calculate the molecular formula.

Determine the empirical formula:

Find the number of moles of each element:

Moles of chromium = 0.70 g / 52.00 g/mol = 0.0135 mol

Moles of sulfur = 0.65 g / 32.06 g/mol = 0.0203 mol

Moles of oxygen = (2.65 g - 0.70 g - 0.65 g) / 16.00 g/mol = 0.0681 mol

Divide each mole value by the smallest one (0.0135 mol) to obtain the empirical formula:

Chromium: 0.0135 mol / 0.0135 mol = 1

Sulfur: 0.0203 mol / 0.0135 mol = 1.50 (rounded to 2)

Oxygen: 0.0681 mol / 0.0135 mol = 5 (rounded to 4)

The empirical formula is CrSO4.

Calculate the molecular formula:

Determine the empirical formula mass:

CrSO4 = 52.00 g/mol + 32.06 g/mol x 2 + 16.00 g/mol x 4 = 156.12 g/mol

Divide the molar mass by the empirical formula mass to obtain the multiplying factor:

392.21 g/mol / 156.12 g/mol = 2.51 (rounded to 3)

Multiply each subscript in the empirical formula by 3 to obtain the molecular formula:

Cr3S3O12

Therefore, the molecular formula of the compound is Cr3S3O12

User Hbirjand
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