Question

a compound contains 40.0% c, 6.71% h, and 53.29% o by mass. the molecular weight of the compound is 60.05 g/mol. the molecular formula of this compound is

ch2o

c2h4o2

c2h2o4

cho2

c2h3o4

question 11
1 pts
a compound that is composed of carbon, hydrogen, and oxygen contains 70.6% c, 5.9% h, and 23.5% o by mass. the molecular weight of the compound is 136 amu. what is the molecular formula?

c8h4o

c9h12o

c4h4o

c8h8o2

c5h6o2

Explanation:

Step1: Find the empirical formula

Assume we have a 100 - g sample of the compound. Then the masses of \(C\), \(H\), and \(O\) are \(m(C)=40.0\ g\), \(m(H) = 6.71\ g\), \(m(O)=53.29\ g\).
The molar masses are \(M(C)=12.01\ g/mol\), \(M(H)=1.008\ g/mol\), \(M(O) = 16.00\ g/mol\).
The number of moles:
\(n(C)=\frac{m(C)}{M(C)}=\frac{40.0\ g}{12.01\ g/mol}\approx3.33\ mol\)
\(n(H)=\frac{m(H)}{M(H)}=\frac{6.71\ g}{1.008\ g/mol}\approx6.66\ mol\)
\(n(O)=\frac{m(O)}{M(O)}=\frac{53.29\ g}{16.00\ g/mol}\approx3.33\ mol\)
Divide by the smallest number of moles (\(n = 3.33\)):
\(C:\frac{3.33}{3.33}=1\), \(H:\frac{6.66}{3.33}=2\), \(O:\frac{3.33}{3.33}=1\)
The empirical formula is \(CH_{2}O\), and its formula mass \(M_{empirical}=(12.01 + 2\times1.008+16.00)\ g/mol=30.03\ g/mol\)

Step2: Find the molecular formula

The molecular formula mass \(M_{molecular}=60.05\ g/mol\)
The ratio \(n=\frac{M_{molecular}}{M_{empirical}}=\frac{60.05\ g/mol}{30.03\ g/mol}\approx2\)
Multiply the empirical formula by \(n\): \((CH_{2}O)\times2 = C_{2}H_{4}O_{2}\)

Answer:

\(C_{2}H_{4}O_{2}\)