1. A 9.941 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 14.57 grams of CO2 an

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1. A 9.941 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 14.57 grams of CO2 an

d 5.966 grams of H2O are produced.
In a separate experiment, the molar mass is found to be 60.05 g/mol.
Determine the empirical formula and the molecular formula of the organic compound.
Enter the elements in the order C, H, O
empirical formula =
molecular formula =

2.A 14.16 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 17.97 grams of CO2 and 4.905 grams of H2O are produced.

In a separate experiment, the molar mass is found to be 104.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Enter the elements in the order C, H, O

empirical formula =

molecular formula =

Chemistry

1 answer:

salantis [7] · Answer 1 · 2022-02-05T12:27:21+03:00

Answer:

1) empirical formula = CH2O ; the molecular formula = C2H402

2) empirical formule = C3H404 = Molecular formula

Explanation:

CxHyOz → CO2 + H2O

⇒ So for CO2 we can say that for each 1 C-atom (12g) there are 2 O-atoms (32g).

This means the ratio is 12g C/ 32g O

CO2 has a molar mass of 44 g/mole

in 14.57 grams of CO2 there is: (12g Carbon / 44g CO2) * 14.57 g CO2 = 3.9736 g Carbon

⇒For H2O we can do the same:

The ratio of each element in H2O is: 2 H-atoms/ 1 O-atom This is 2g H/ 16g O

In 5.966 g of water there are: [2 g H / 18 g H2O] * 5.966 g H2O = 0.6629 g H

⇒The original sample had 9.941 g of Sample - 3.9736 g of C - 0.6629 g of H = 5.3045 g of O

2) Calculate number of moles

C: 3.9736 g / 12.0 g/mol = 0.3311 mol

H: 0.6629 g / 1.0 g/mol = 0.6629 mol

O: 5.3045g / 16.0 g/mol = 0.3315 mol

3) Now we should divide each number of mole by the smallest number (0.3311) to find the proportion of the elements

C: 0.3311/ 0.3311= 1

H: 0.6629 / 0.3311 = 2

O: 0.3315/ 0.3311 = 1

This gives us the empirical formula of CH2O

4) Calculate the mass of the empirical formula

Molar mass of Carbon = 12g /mole

Molar mass of Hydrogen = 1g /mole

Molar mass of Oxygen = 16g /mole

mass of the empirical formule = 12 + 2*1 + 16 = 30g

5) Calculate the molecular formula

mass of molecular formule / mass of empirical formula = n

We have to multiply the empirical formula by n to get the molecular formula.

60 / 30 = 2 = n

If we multiply CH2O by 2 we'll get: C2H4O2

If we control this by calculating the molar mass:

2*12 + 4*1.01 + 2*16 = 60.04 g/mole

Then, the molecular formula is C2H4O2

CxHyOz → CO2 + H2O

⇒ So for CO2 we can say that for each 1 C-atom (12g) there are 2 O-atoms (32g).

This means the ratio is 12g C/ 32g O

CO2 has a molar mass of 44 g/mole

in 17.97 grams of CO2 there is: (12g Carbon / 44g CO2) * 17.97 g CO2 = 4.90 g Carbon

⇒For H2O we can do the same:

The ratio of each element in H2O is: 2 H-atoms/ 1 O-atom This is 2g H/ 16g O

In 4.905 g of water there are: [2 g H / 18 g H2O] * 4.905 g H2O = 0.545 g H

⇒The original sample had 14.16 g of Sample - 4.90 g of C - 0.545 g of H = 8.715g of O

2) Calculate number of moles

C: 4.90 g / 12.0 g/mol = 0.4083 mol

H: 0.545 g / 1.0 g/mol = 0.545 mol

O: 8.715g / 16.0 g/mol = 0.5447 mol

3) Now we should divide each number of mole by the smallest number (0.4083) to find the proportion of the elements

C: 0.4083/ 0.4083= 1

H: 0.545 / 0.4083 =1.33

O: 0.5447/ 0.4083 = 1.33

We should multiply everything by 3

This gives us the empirical formula of C3H4O4