Question

During an analysis of 255 mg of a molecule that contains only C, H and O. As a result 561 mg CO2 is
produced along with 306 mg H2O
(a) If the molecule contains only C, H and O, what is empirical formula?
(b) If the molar mass of the compound is 180 g/mol, what is molecular formula of the compound

Answer 1

Answer:

153

C

9

H

24

O

3

C

x

H

y

O

z

on combustion we have :

C

x

H

y

O

z

+

(

x

+

y

4

−

z

2

)

O

2

→

x

C

O

2

+

y

2

H

2

O

moles of

C

are equal to moles of

C

O

2

so :

n

C

O

2

=

561

44

=

12.75

×

10

−

3

n

C

=

n

C

O

2

=

12.75

×

10

−

3

∴

w

C

=

12

×

12.75

=

153

mg

moles of

H

are equal to twice of moles of

H

2

O

so :

n

H

=

2

×

n

H

2

O

=

2

×

306

18

×

10

−

3

=

34

×

10

−

3

∴

w

H

=

34

mg

w

C

+

w

H

=

153

+

34

=

187

mg

w

O

=

w

compound

−

187

=

255

−

187

=

68

mg

so moles of

O

:

n

O

=

68

16

=

4.25

×

10

−

3

=

4.25

mmol

therefore the ratio

C

:

H

:

O

=

12.75

:

34

:

4.25

=

3

:

8

:

1

therefore Empirical formula :

C

3

H

8

O

Empirical mass

=

60

Empirical mass

=

molecular mass

n

(molar mass of compound

=

180

g/mol

and n =integer)

so we get

n

=

3

Emprical formula

=

molecular formula

n

so, Molecular formula

=

C

9

H

24

O

3

Answer 2

The produced along with molecular mass $C_{9}H_{24}O_{3}$.

Explanation:

moles of $C$ are equal to moles of $CO_{2}$

$nco_{2}=\frac{561}{44}$

$=12.75*10^{-3}$

$nc=nco_{2}=12.75*10^{-3}$

$w_{c}=12*12.75$

$=153mg$

moles of $H$ are equal to twice of moles of $H_{2}O$

$n_{H}=2*n_{H}_{2}o$

$=2*\frac{306}{18}*10^{-3}$

$=34*10^{-3}$

$w_{H}=34mg$

$w_{C}+w_{H}=153+34$

$=187mg$

$w_{O}=W_{compound}-187$

$=255-187$

$=68mg$

$n_{O}=\frac{68}{16}$

$=4.25*10^{-3}$

$=4.25mmol$

$C:H:O=12.75:34:4.25$

$=3:8:1$

Empirical formula:$C_{3}H_{8}O$

Empirical mass $=60$

Empirical mass$=\frac{molecularformula}{n}$

$n=3$

molecular formula=$C_{9}H_{24}O_{3}$.