Question

1. What is the total sample size (in grams) for a sample of aluminum acetate which contains 2.71 g of oxygen?

2. What is the total sample size (in grams) for a sample of ethanol (CH3CH2OH) which contains 2.98 g of oxygen?

3. What is the total sample size (in grams) for a sample of sodium nitrite which contains 3.08 g of oxygen?

Answer 1

1)

aluminium acetate formula is Al(C2H3O2)3

Molar mass of O = 16 g/mol

mass(O)= 2.71 g

number of mol of O,

n = mass of O/molar mass of O

=(2.71 g)/(16 g/mol)

= 0.1694 mol

1 mol of Al(C2H3O2)3 contains 6 moles of O

So,

moles of Al(C2H3O2)3 = moles of O / 6

= 0.1694/6

= 2.823*10^-2 mol

Molar mass of Al(C2H3O2)3,

MM = 1*MM(Al) + 6*MM(C) + 9*MM(H) + 6*MM(O)

= 1*26.98 + 6*12.01 + 9*1.008 + 6*16.0

= 204.112 g/mol

mass of Al(C2H3O2)3,

m = number of mol * molar mass

= 2.823*10^-2 mol * 204.112 g/mol

= 5.762 g

Answer: 5.76 g

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