A 0.338-g sample of antimony was completely reacted with 0.295 g of chlorine gas

Question

A 0.338-g sample of antimony was completely reacted with 0.295 g of chlorine gas to form antimony chloride. Determine the empirical formula of the antimony chloride. A 0.058-g sample of copper was reacted with oxygen to form 1.074 g of copper oxide. Determine the empirical formula of the copper oxide. A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2.156 g of pure zinc metal. Determine the empirical formula of the initial zinc oxide. A 0.700-g sample of phosphorous is reacted with excess oxygen to form a product with a mass of 1.604 g. Determine the empirical formula of the product. If product is known have a molecular mass of approximately 284 g/mol, what is the molecular formula the product? (Show all calculations required to arrive at the answers.) empirical formula:molecular formula:

Explanation / Answer

1. moles Sb = 0.338 g/121.76 g/mol = 2.776 mmol

moles Cl2 = 0.295 g/71 g/mol = 4.155 mmol

moles Cl2/moles Sb = 4.155/2.776 = 1.5

So emiprical formula = 1.5(SbCl2) = SbCl3

2. moles Cu = 0.858 g/63.546 g/mol = 0.0135 mol

moles CuO = moles O = 1.074 g/79.545 g/mol = 0.0135 mol

ratio moles (O/Cu) = 1

Empirical formula = CuO

3. moles Zn formed = 2.156 g/65.38 g/mol = 0.033 mol

moles ZnO = 2.684 g/81.408 g/mol = 0.033 mol

moles ZnO = moles Zn = 0.033 mol

empirical formula = ZnO

4. moles P = 0.7 g/31 g/mol = 0.0225 mol

moles O = (1.604 - 0.7)/16 = 0.0565 mol

divide by smallest factor,

P = 0.0225/0.0225 = 1

O = 0.0565/0.0225 = 2.5

Empirical formula = PO2.5

mass ratio = (284/(31 + 2.5 x 16)) = 4

molecular formula = 4(PO2.5) = P4O10

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