Question

a compound is found to be 25.53% mg and 74.47% cl by mass. its molar mass is 95.21g/mol. what is the molecular formula?

Explanation:

Step1: Assume 100g of the compound

If we assume 100g of the compound, then the mass of \(Mg = 25.53g\) and the mass of \(Cl=74.47g\).

Step2: Calculate the number of moles of each element

The molar mass of \(Mg\) is \(M_{Mg}=24.31g/mol\), and the molar mass of \(Cl\) is \(M_{Cl} = 35.45g/mol\).
The number of moles of \(Mg\), \(n_{Mg}=\frac{m_{Mg}}{M_{Mg}}=\frac{25.53g}{24.31g/mol}\approx1.05mol\)
The number of moles of \(Cl\), \(n_{Cl}=\frac{m_{Cl}}{M_{Cl}}=\frac{74.47g}{35.45g/mol}\approx2.1mol\)

Step3: Find the ratio of moles

Divide each number of moles by the smaller number of moles (\(n_{Mg}\approx1.05mol\))
For \(Mg\): \(\frac{n_{Mg}}{n_{Mg}} = 1\)
For \(Cl\): \(\frac{n_{Cl}}{n_{Mg}}=\frac{2.1mol}{1.05mol}=2\)
The empirical formula is \(MgCl_{2}\)

Step4: Calculate the empirical formula mass

The empirical formula mass of \(MgCl_{2}\), \(M_{empirical}=(24.31 + 2\times35.45)g/mol=(24.31+70.9)g/mol = 95.21g/mol\)

Step5: Determine the molecular formula

The molar mass of the compound \(M = 95.21g/mol\)
Since \(\frac{M}{M_{empirical}}=\frac{95.21g/mol}{95.21g/mol}=1\)

Answer:

The molecular formula is \(MgCl_{2}\)