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An unknown compound contains only the three elements C, H, and O. A pure sample of the compound is analyzed and found to be 65.60 percent C and 9.44 percent H by mass. (a) Determine the empirical formula of the compound.

User Francois G
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2 Answers

6 votes

Final answer:

To determine the empirical formula, convert the mass percentages to grams for a 100 g sample, calculate moles of each element, find the simplest whole number mole ratio, and derive the empirical formula accordingly. For a compound with 65.60% C and 9.44% H, the empirical formula would be C7H12O2.

Step-by-step explanation:

To determine the empirical formula of a compound with given percentages of carbon (C), hydrogen (H), and oxygen (O), we convert these percentages into grams as if we have a 100 g sample. This way, the percent by mass is numerically equal to the mass of each element in grams. Afterward, we convert these masses to moles by dividing by the atomic mass of each element. Next, we find the simplest whole number ratio of the moles of each element to each other, which may involve dividing all the moles by the smallest number of moles measured. This ratio provides us with the empirical formula.

For example, if a compound contains 65.60% C and 9.44% H, since oxygen is the only other element, it must make up the remaining 24.96% by mass. If we assume a 100 g sample, we have 65.60 g C, 9.44 g H, and 24.96 g O. Dividing these values by the respective atomic masses (12.01 g/mol for C, 1.008 g/mol for H, and 16.00 g/mol for O), we get 5.46 mol C, 9.37 mol H, and 1.56 mol O. Dividing by the smallest number of moles (1.56 mol for O), we get approximately 3.5 mol C, 6 mol H, and 1 mol O. To get whole numbers, we can multiply each by 2 to get the empirical formula C7H12O2.

User Bigpotato
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Answer: The empirical formula is
C_7H_(12)O_2

Step-by-step explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 65.60 g

Mas of H = 9.44

Mass of O = (100-(65.6+9.44)= 24.96 g

Step 1 : convert given masses into moles.

Mass of C =
\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (65.5g)/(12g/mole)=5.47moles

Moles of H =
\frac{\text{ given mass of N}}{\text{ molar mass of N}}= (9.44g)/(1g/mole)=9.44moles

Moles of O=
\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (24.96g)/(16g/mole)=1.56moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =
(5.47)/(1.56)=3.5

For H =
(9.44)/(1.56)=6

For O =
(1.56)/(1.56)=1

The ratio of C:H:O= 3.5: 6: 1

Converting them into whole numbers, the ratio will be 7: 12: 2

Hence the empirical formula is
C_7H_(12)O_2

User Sqoo
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