Answer:
C4H3O
Explanation:
Data obtained from the question include:
Carbon (C) = 68.85%
Hydrogen (H) = 4.95%
Since the vaccine contains carbon, hydrogen and oxygen, the remaining percentage will be for oxygen. The percentage of oxygen will be:
Oxygen (O) = 100 - (68.85 + 4.95)
= 26.2%
The empirical formula for the vaccine can be obtained as follow:
Carbon (C) = 68.85%
Hydrogen (H) = 4.95%
Oxygen (O) = 26.2%
Divide the above by their individual molar mass as shown below:
C = 68.85/12 = 5.7375
H = 4.95/1 = 4.95
O = 26.2/16 = 1.6375
Next, divide by the smallest number as shown below:
C = 5.7375/1.6375 = 4
H = 4.95/1.6375 = 3
O = 1.6375/1.6375 = 1
Therefore, the empirical formula for the vaccine is C4H3O
Answer:
[tex]C_7H_6O_2[/tex]
Explanation:
We have to assume that we have 100 g of the unknow as first step. With this in mind we can calculate the grams of C, H and O in the sample.
[tex]100~g~\frac{68.85}{100}=68.85~g~of~C[/tex]
[tex]100~g~\frac{4.95}{100}=4.95~g~of~H[/tex]
For the calculation of O we have to know the percentage of O therefore we can add the percentages of H and C and then do a substraction from 100 %, so:
[tex]%~of~O=~100-(68.85+4.95)=~26.2~%~of~O[/tex]100-(68.85+4.95)=26.2 %
With this value we can calculate the amount of O in grams in the sample:
[tex]100~g~\frac{26.2}{100}=26.2~g~of~O[/tex]
The next step is the calculation of moles for each atom, for this we have to know the atomic mass of each atom ( C: 12 g/mol; H 1 g/mol; O 16 g/mol):
[tex]68.85~g~of~C\frac{1~mol~C}{12~g~C}=~5.74~mol~C[/tex]
[tex]4.95~g~of~H\frac{1~mol~H}{1~g~H}=4.95~mol~of~H[/tex]
[tex]26.2~g~of~O\frac{1~mol~O}{16~g~O}=1.64~mol~of~O[/tex]
Now we have to divide by the smallest value, so:
[tex]\frac{5.74~mol~C}{1.64}=3.5[/tex]
[tex]\frac{4.95~mol~H}{1.64}=3[/tex]
[tex]\frac{1.64~mol~O}{1.64}=1[/tex]
We obtain an decimal number for C, therefore we have to multiply all by "2", so:
[tex]3.5x2= 7~mol~C[/tex]
[tex]3x2= 6~mol~H[/tex]
[tex]1x2= 2~mol~O[/tex]
Therefore the formula would be: [tex]C_7H_6O_2[/tex]
