Answer:
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.
