Verified Answer :
empirical formula is the simplest ratio of whole numbers of components making up a compound
the percentage composition of each element has been given so we are going to find the mass for 100 g of the compound
mass C - 59.0 g
H - 7.1 g
O - 26.2 g
N - 7.7g
number of moles
C - 59.0 g/ 12 g/mol
H - 7.1 g/ 1 g/mol
O - 26.2 g/ 16 g/mol
N - 7.7 g / 14 g/mol
C = 4.9 mol H= 7.1 mol O= 1.6 mol N =0.55 mol
divide by the least number of moles
C - 4.9 / 0.55 = 8.9
H - 7.1 / 0.55 = 12.9
O -1.6 / 0.55 = 2.9
N - 0.55/ 0.55 = 1.0
the number of atoms of each element rounded off
C - 9
H - 13
O - 29
N - 1
the empirical formula is - C₉H₁₃O₃N
Answer : The empirical formula of a compound is, \(C_{9}H_{13}O_3N\)
Solution : Given,
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C = 59 g
Mass of H = 7.1 g
Mass of O = 26.2 g
Mass of N = 7.7 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of O = 16 g/mole
Molar mass of N = 14 g/mole
Step 1 : convert given masses into moles.
Moles of C = \(\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{59g}{12g/mole}=4.9moles\)
Moles of H = \(\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.1g}{1g/mole}=7.1moles\)
Moles of O = \(\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{26.2g}{16g/mole}=1.6moles\)
Moles of N = \(\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{7.7g}{14g/mole}=0.55moles\)
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = \(\frac{4.9}{0.55}=8.9\approx 9\)
For H = \(\frac{7.1}{0.5}=12.9\approx 13\)
For O = \(\frac{1.6}{0.5}=2.9\approx 3\)
For N = \(\frac{0.55}{0.55}=1\)
The ratio of C : H : O : N = 9 : 13 : 3 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = \(C_{9}H_{13}O_3N_1\) = \(C_{9}H_{13}O_3N\)
Therefore, the empirical formula of a compound is, \(C_{9}H_{13}O_3N\)