The empirical formula of the compound is C₄H₅O₃.
Explanation:
If hydrogen % = 5.98823, the Sum of C+O = 94.01177% or 47.005885% each
The original mass of 300 g is not important
C = 47.005885 / 12.011 = 3.913569
H = 5.98823 / 1.008 = 5.94070
O = 47.005885 / 15.999 = 2.93805
Divide by the smallest number:
C =1.33
H = 2.022
O = 1
Multiply through by 3
C = 4
H =6
O = 3
The empirical formula of a compound = C₄H₆O₃
Do not be confused by the 300 g. It is totally irrelevant to the question because you are dealing with % amounts. It would only have been of importance if you were given some mass values.
Considering the definition of empirical formula, the empirical formula is H₆C₄O₃.
The empirical formula is the simplest expression to represent a chemical compound, which indicates the elements that are present and the minimum proportion in whole numbers that exist between its atoms, that is, the subscripts of chemical formulas are reduced to the most integers. small as possible.
In this case, the compound contains only carbon, hydrogen and oxygen. The compound contains the exact same percentage of carbon as it has oxygen. The percentage of hydrogen is known to be 5.98823%.
Then, the sum of percentage of carbon and oxygen is 94.01177%. Then, the percentage of carbon is 47.005885%, and the percentage of oxygen is 47.005885%.
Assuming a 100 grams sample, the percentages match the grams in the sample. So you have 5.98823 grams of hydrogen, 47.005885 grams of carbon and 47.005885 grams of oxygen.
Then it is possible to calculate the number of moles of each atom in the molecule, taking into account the corresponding molar mass:
Hydrogen: [tex]\frac{5.98823 grams}{1.008 \frac{grams}{mole} }=[/tex] 5.94070 moles
Carbon: [tex]\frac{47.005885 grams}{12.011 \frac{grams}{mole} }=[/tex]3.913569 moles
Oxygen: [tex]\frac{47.005885 grams}{16 \frac{grams}{mole} }=[/tex]2.93786 moles
The empirical formula must be expressed using whole number relationships, for this the numbers of moles are divided by the smallest result of those obtained. In this case:
Hydrogen: [tex]\frac{5.94070 moles}{2.93786 moles}=[/tex] 2.022
Carbon: [tex]\frac{3.913569 moles}{2.93786 moles}=[/tex] 1.33
Oxygen: [tex]\frac{2.93786 moles}{2.93786 moles}=[/tex]1
As in the empirical formula the numbers must be expressed in whole numbers, the ratio of atoms is multiplied by some number with which everything is obtained in simple whole numbers (in this case by 3).
Hydrogen: 2.022×3= 6.066 ≅ 6
Carbon: 1.33× 3= 3.99 ≅4
Oxygen: 1× 3= 3
Therefore the H: C: O mole ratio is 6: 4: 3
Finally, the empirical formula is H₆C₄O₃.
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