Question

1. what is the empirical formula of a compound that is 64.8% c, 13.62% h, and 21.58% o?
2. a. find the empirical formula for the compound that contains 42.05 g of nitrogen and 95.95 g of oxygen.

b. if the molecular mass of this compound is 138 g/mole what is the molecular formula?

Explanation:

1. What is the empirical formula of a compound that is \(64.8\%\text{ C}\), \(13.62\%\text{ H}\), and \(21.58\%\text{ O}\)?

Step1: Assume 100g of the compound

If we assume we have 100g of the compound, then the masses of \(C\), \(H\), and \(O\) are \(m_{C}=64.8g\), \(m_{H} = 13.62g\), \(m_{O}=21.58g\)

Step2: Calculate the number of moles of each element

The molar mass of \(C\) is \(M_{C}=12.01g/mol\), the molar mass of \(H\) is \(M_{H}=1.008g/mol\), and the molar mass of \(O\) is \(M_{O} = 16.00g/mol\)
The number of moles of \(C\): \(n_{C}=\frac{m_{C}}{M_{C}}=\frac{64.8g}{12.01g/mol}\approx5.40mol\)
The number of moles of \(H\): \(n_{H}=\frac{m_{H}}{M_{H}}=\frac{13.62g}{1.008g/mol}\approx13.51mol\)
The number of moles of \(O\): \(n_{O}=\frac{m_{O}}{M_{O}}=\frac{21.58g}{16.00g/mol}\approx1.35mol\)

Step3: Divide by the smallest number of moles

Divide the number of moles of each element by the smallest number of moles (\(n_{O} = 1.35mol\))
For \(C\): \(\frac{n_{C}}{n_{O}}=\frac{5.40mol}{1.35mol}=4\)
For \(H\): \(\frac{n_{H}}{n_{O}}=\frac{13.51mol}{1.35mol}\approx10\)
For \(O\): \(\frac{n_{O}}{n_{O}} = 1\)

Step1: Calculate the number of moles of each element

The molar mass of \(N\) is \(M_{N}=14.01g/mol\), the molar mass of \(O\) is \(M_{O}=16.00g/mol\)
The number of moles of \(N\): \(n_{N}=\frac{m_{N}}{M_{N}}=\frac{42.05g}{14.01g/mol}\approx3.00mol\)
The number of moles of \(O\): \(n_{O}=\frac{m_{O}}{M_{O}}=\frac{95.95g}{16.00g/mol}\approx6.00mol\)

Step2: Divide by the smallest number of moles

Divide the number of moles of each element by the smallest number of moles (\(n_{N}=3.00mol\))
For \(N\): \(\frac{n_{N}}{n_{N}} = 1\)
For \(O\): \(\frac{n_{O}}{n_{N}}=\frac{6.00mol}{3.00mol}=2\)

Step1: Calculate the empirical formula mass

The empirical formula is \(NO_{2}\). The empirical formula mass \(M_{EF}=(14.01 + 2\times16.00)g/mol=(14.01+32.00)g/mol = 46.01g/mol\)

Step2: Calculate the ratio \(n\)

\(n=\frac{M_{MF}}{M_{EF}}\), where \(M_{MF} = 138g/mol\) and \(M_{EF}=46.01g/mol\)
\(n=\frac{138g/mol}{46.01g/mol}\approx3\)

Step3: Determine the molecular formula

Multiply the sub - scripts in the empirical formula by \(n\). The empirical formula is \(NO_{2}\), so the molecular formula is \((NO_{2})\times3=N_{3}O_{6}\)

Answer:

The empirical formula is \(C_{4}H_{10}O\)

2. a. Find the empirical formula for the compound that contains \(42.05g\) of nitrogen and \(95.95g\) of oxygen