Question

an addictive substance being analyzed in the lab has a molar mass of 162.23 g/mol. a 100 gram sample contains 74.0% carbon, 8.7% hydrogen, and 17.3% nitrogen. what would this substances empirical formula be?
empirical formula
weight = ? amu
c h n
5 7 1
molecular formula
weight = 162 amu
c h n
molecular formula weight
composition by % mass
relative mass percent or mass (in grams) of each element in the substance
moles of each element in the substance

Explanation:

Step1: Calculate moles of each element

Assume we have a 100 - gram sample.

• Moles of \(C\): \(n_{C}=\frac{74.0\ g}{12.01\ g/mol}\approx6.16\ mol\)
• Moles of \(H\): \(n_{H}=\frac{8.7\ g}{1.008\ g/mol}\approx8.63\ mol\)
• Moles of \(N\): \(n_{N}=\frac{17.3\ g}{14.01\ g/mol}\approx1.23\ mol\)

Step2: Divide by the smallest number of moles

Divide each mole value by \(n_{N} = 1.23\ mol\)

• For \(C\): \(\frac{6.16}{1.23}\approx5\)
• For \(H\): \(\frac{8.63}{1.23}\approx7\)
• For \(N\): \(\frac{1.23}{1.23}=1\)

Answer:

The empirical formula is \(C_{5}H_{7}N\)