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xenn [34]
3 years ago
15

An unknown compound contains only the three elements C, H, and O. A pure sample of the compound is analyzed and found to be 65.6

0 percent C and 9.44 percent H by mass. (a) Determine the empirical formula of the compound.
Chemistry
1 answer:
Bas_tet [7]3 years ago
4 0

Answer: The empirical formula is C_7H_{12}O_2

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}}= \frac{65.5g}{12g/mole}=5.47moles

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

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{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 = \frac{5.47}{1.56}=3.5

For H = \frac{9.44}{1.56}=6

For O =\frac{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

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