Vitamin C is essential for the prevention of scurvy. Combustion of 0.2000g of vitamin
C gives 0.2998g of CO2 and 0.819g of H20. What is the empirical formula of vitamin C?
(please show the steps also)
Answers
moles CO2 = 0.2998 g CO2 x 1 mole/44 g = 0.006814 moles CO2
mole C = 0.006814 moles CO2 x 1 mole C/mole CO2 = 0.006814 moles C
moles H2O = 0.0819 g H2O x 1 mole/18 g = 0.00455 moles H2O
moles H = 0.00455 moles H2O x 2 moles H/mole H2O = 0.0091 moles H
In order to find mass of oxygen (O) in the compound, we need to first convert the moles of C and mole of H to grams.
0.006814 moles C x 12 g/1 mole = 0.08177 g Carbon
0.0091 moles H x 1 g/1 mole = 0.0091 g Hydrogen
Total mass of C and H = 0.08177 g + 0.0091 g = 0.09087 g
Grams of O = 0.2000 g - 0.09087 g = 0.1091 g
moles O = 0.1091 g O x 1 mole/16 g = 0.006818 moles O
Now, to get the empirical formula, use moles C, moles H and moles O
C = 0.00681
H = 0.0091
O = 0.00681
Divide all by the smallest, i.e. 0.00681 to get ...
C = 1.00
H = 1.33
O = 1.00
Now, multiply all by 3 to get whole numbers, and the answer is...
C3H4O3 = empirical formula for ascorbic acid
Answer:
Explanation:
given: molar mass = 0.2g
mass of = 0.2998g
mass of = 0.819g
molecular mass of CO2 = 12 + 2*(16) = 44g
amount of C = 12/44 * 0.2998
= 0.0818g
molecular mass of H2O = 2+16=18g
amount of H = 1*2/18 * 0.819
= 0.0091g
[NOTE: we are not finding O along with this because O is present in CO2 and H2O]
mass% = mass of element in compound*100/molar mass of compound
mass% C = 0.0818/0.2 *100 =40.9%
mass% H= 0.0091/0.2 *100 = 4.55%
mass% O = 100-(40.9+4.55) = 54.55%
[calculate the number of moles]
no of moles of C = 40.9/12 = 3.408
no of moles of H = 4.55/1 = 4.55
no of moles of O = 54.55=16 = 3.41
[divide by the smallest value among these i.e 3.41]
3.408/3.41 = 1
4.55/3.41=1.33
3.41/3.41 =1
[multiply by a common number to convert it into whole numbers]
multiplying by 3 we get, 3,4,3 respectively
therefore, vitamin X =
Hope this helps :)