Question

a compound is analyzed as 17.6% na, 39.7% cr, and 42.7% o. what is the empirical formula of the substance? nacro₃ na₂cr₂o₇ na₂cro₄ na₂cr₂o₆

Explanation:

Step1: Assume 100g of the compound

Assume we have 100g of the compound. Then the masses are \(m_{Na}=17.6g\), \(m_{Cr}=39.7g\), \(m_{O}=42.7g\)

Step2: Calculate the number of moles

The molar mass of \(Na\) is \(M_{Na} = 22.99g/mol\), the molar mass of \(Cr\) is \(M_{Cr}=52.00g/mol\), and the molar mass of \(O\) is \(M_{O} = 16.00g/mol\)

The number of moles of \(Na\): \(n_{Na}=\frac{m_{Na}}{M_{Na}}=\frac{17.6g}{22.99g/mol}\approx0.765mol\)

The number of moles of \(Cr\): \(n_{Cr}=\frac{m_{Cr}}{M_{Cr}}=\frac{39.7g}{52.00g/mol}\approx0.763mol\)

The number of moles of \(O\): \(n_{O}=\frac{m_{O}}{M_{O}}=\frac{42.7g}{16.00g/mol}\approx2.67mol\)

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (\(n = 0.763mol\))

For \(Na\): \(\frac{n_{Na}}{n}=\frac{0.765mol}{0.763mol}\approx1\)

For \(Cr\): \(\frac{n_{Cr}}{n}=\frac{0.763mol}{0.763mol} = 1\)

For \(O\): \(\frac{n_{O}}{n}=\frac{2.67mol}{0.763mol}\approx3.5\)

Multiply each ratio by 2 to get whole - numbers. So \(Na:Cr:O = 2:2:7\)

Answer:

\(Na_{2}Cr_{2}O_{7}\)