Question

part c ( 25 marks) - show all the steps and working in the answers for full credit. also, write proper units for your answer.

1. a compound is 38.62% na 26.9% s and 34.48% o. its molar mass is 234g/mol.

find its empirical formula and molecular formula.

Explanation:

Step1: Calculate moles of each element

Assume we have 100g of the compound. Then mass of Na = 38.62g, mass of S = 25.9g, mass of O = \(100-(38.62 + 25.9)=35.48\)g.
Molar mass of Na (\(M_{Na}\)) = 23g/mol, moles of Na (\(n_{Na}\))=\(\frac{38.62}{23}=1.679\)mol.
Molar mass of S (\(M_{S}\)) = 32g/mol, moles of S (\(n_{S}\))=\(\frac{25.9}{32}=0.809\)mol.
Molar mass of O (\(M_{O}\)) = 16g/mol, moles of O (\(n_{O}\))=\(\frac{35.48}{16}=2.2175\)mol.

Step2: Find the mole - ratio

Divide each mole value by the smallest mole value (\(n_{S} = 0.809\)mol)
For Na: \(\frac{1.679}{0.809}\approx2\)
For S: \(\frac{0.809}{0.809} = 1\)
For O: \(\frac{2.2175}{0.809}\approx2.75\). Multiply by 4 to get whole - numbers.
Na: \(2\times4 = 8\), S: \(1\times4=4\), O: \(2.75\times4 = 11\)
The empirical formula is \(Na_{2}SO_{4}\) (after simplifying the ratio \(8:4:11\) to \(2:1:4\) by dividing by 4).

Step3: Calculate the empirical formula mass

Empirical formula mass of \(Na_{2}SO_{4}\): \(2\times23+32 + 4\times16=46 + 32+64 = 142\)g/mol.

Step4: Find the multiple (\(n\))

Given molar mass \(M = 284\)g/mol. \(n=\frac{M}{M_{empirical}}=\frac{284}{142}=2\)

Step5: Determine the molecular formula

Multiply the sub - scripts in the empirical formula by \(n\). Molecular formula is \(Na_{4}S_{2}O_{8}\)

Answer:

Empirical formula: \(Na_{2}SO_{4}\), Molecular formula: \(Na_{4}S_{2}O_{8}\)