Question

You perform a combustion analysis on a 255 mg sample of a substance that contains only C, H, and O, and you find that 561 mg of CO2 is produced, along with 306 mm of H2O.
if the substance contains only C, H, and O, what is the empirical formula
if the molar mass of the compound is 180 g/mol what is the molecular formula of the compound

Answer 1

Answer:

molecular formula of compound =C₉H₂₄O₃

Explanation:

The compound contains only C,H and O

The combustion equation for carbon is:

$C+O_{2}-->CO_{2}$

Thus if we are obtaining 44g of carbon dioxide it means the amount of carbon is 12g.

If we are getting 1g of carbon dioxide, it means the amount of carbon = $\frac{12}{44}g$

If we are getting 561mg of carbon dioxide, it means the amount of carbon=$\frac{12X561}{44}=153mg$

So the mass of carbon in the given sample = 153mg

Percentage of carbon = $\frac{massofcarbonX100}{massofsample}=\frac{153X100}{255}=60$

The combustion equation for hydrogen is:

$H_{2}+0.5O_{2}-->H_{2}O$

Thus if we are obtaining 18g of water it means the amount of hydrogen is 2g.

If we are getting 1g of water, it means the amount of hydrogen = $\frac{2}{18}g$

If we are getting 306mg of water, it means the amount of hydrogen=$\frac{2X306}{18}=34mg$

So the mass of hydrogen in the given sample = 34mg

Percentage of hydrogen = $\frac{massofhydrogenX100}{massofsample}=\frac{34X100}{255}=13.33$

Percentage of oxygen=100-(60+13.33)=26.67%

The moles of each element in 100g of sample will be

moles of carbon = $\frac{mass}{atomicmass}=\frac{60}{12}=5$

moles of hydrogen = $\frac{mass}{atomicmass}=\frac{13.33}{1}=13.33$

moles of oxygen = $\frac{mass}{atomicmass}=\frac{26.67}{16}=1.67$

Let us divide all the moles with 1.67

moles of carbon = 3

moles of hydrogen = 8

moles of oxygen = 1

So the empirical formula of the compound would be:C₃H₈O

The empirical mass = 60

molar mass = 180g

So the molecular formula will be = $\frac{180}{60}Xempiricalformula$

molecular formula=C₉H₂₄O₃

Answer 2

The combustion of an organic compound is mostly written as,
                        CaHbOc + O2 --> CO2 + H2O
where a, b, and c are supposed to be the subscripts of the elements C, H, and O in the compound. Determining the number of moles of C and H in the product which is the same as that in the compound,
   (Carbon, C)   :   (561 mg) x (12/44) = 153 mg x (1 mmole/12 mg) = 12.75
   (Hydrogen, H) :  (306 mg) x (2/18)  = 34 mg x (1 mmole/1 mg) = 34 
   Calculating for amount of O in the sample,
(oxygen, O) = 255 - 153 mg - 34 mg = 68 mg x (1mmole/16 mg) = 4.25 
The empirical formula is therefore,
                        C(51/4)H34O17/4
                           C3H8O1
The molar mass of the empirical formula is 60. Therefore, the molecular formula of the compound is,
                                C9H24O3