Question

question 7
1 pts
a compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen,
and 22.83% oxygen. what is the empirical formula?

c7h7o3

c4h6o

c6h11o

c8h13o3

question 8
1 pts
what is the molecular formula of a compound whose empirical formula is ch2o
and molecular weight is 60.0 g/mole?

ch2o

c2h4o2

c12h24o12

c6h12o6

Explanation:

Question 7

Step1: Assume 100g of the compound

If we assume we have 100g of the compound, then the mass of \(C = 68.54g\), mass of \(H=8.63g\), mass of \(O = 22.83g\)

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 \(C\), \(n_{C}=\frac{68.54g}{12.01g/mol}\approx5.71mol\)
The number of moles of \(H\), \(n_{H}=\frac{8.63g}{1.008g/mol}\approx8.56mol\)
The number of moles of \(O\), \(n_{O}=\frac{22.83g}{16.00g/mol}\approx1.43mol\)

Step3: Divide by the smallest number of moles

Divide each number of moles by \(n_{O}\approx1.43mol\)
For \(C\): \(\frac{5.71mol}{1.43mol}\approx4\)
For \(H\): \(\frac{8.56mol}{1.43mol}\approx6\)
For \(O\): \(\frac{1.43mol}{1.43mol} = 1\)

Step1: Calculate the empirical formula weight

The empirical formula \(CH_{2}O\) has an empirical formula weight \(M_{empirical}=(12.01 + 2\times1.008+16.00)g/mol=(12.01 + 2.016 + 16.00)g/mol = 30.026g/mol\)

Step2: Find the ratio \(n\)

We know that the molecular weight \(M_{molecular}=60.0g/mol\)
The ratio \(n=\frac{M_{molecular}}{M_{empirical}}=\frac{60.0g/mol}{30.026g/mol}\approx2\)

Step3: Determine the molecular formula

Multiply each sub - script in the empirical formula by \(n = 2\)
The molecular formula is \(C_{2}H_{4}O_{2}\)

Answer:

\(C_{4}H_{6}O\)

Question 8