Elemental analysis of a sugar molecule called sucrose gives the following mass percent composition: C 40.00%, H 6.72 %, O 53.28 %. The molar mass of sucrose is 342.3g/mol. Find the molecular formula of sucrose. (6 pts)

C12H22O11

Explanation:

1. First you have to chage percentages into grams:

Total molar mass is equal to 100%

100% = 342.3 g

C:

40.00% x (342.3 g/100%) = 136.92 g

H:

6.72%x (342.3 g / 100%) = 23 g

O:

53.28% x (342.3 g/100%) = 182.38 g

2. Then change grams into mol:

C:

1 mol = 12.011 g/m

136.92 g / (12.01ol1 g/mol) = 11.3996 mol

H:

1 mol = 1,008 g/mol

23 g / (1,008 g/mol) = 22.8175 mol

O:

1 mol = 15.999g/mol

182.38 g / (15.999 g/mol) = 11.3995 mol

3. Then divide the values obtained by the smallest value: 11.3995

C:

11.3996 mol / 11.3995 mol = 1

H:

23mol / 11. 3995 mol = 2

O:

11.3995 mol/11.3995 mol = 1

4. Obtein the empirical formula: CH2O

5. Find the epirical formula molar mass:

C = 12.011 g

H = 2 g

O = 16g

CH2O = 12.011 + 2.016+15.999 = 30.026 g

6. Then, divide the molecular formula molar mass by the empirical formula molar mass:

molecular formula molar mass / empirical formula molar mass

342.3 g / 30 g = 11.40

Aproximate the number obteined to a whole number : 11

7. Multiply the subscripst by the whole number 11:

C (1x11) H (2x11) O (1 x 11) = C11 H22 O11

But, the real molecular formula is C12H22O11 because of the carbon who unites the two sugars.