Answer to Question #97050 in General Chemistry for Grace

Question #97050
Determine the empirical formula for the compound C=42.86% H=2.4% N=5.88% Na=9.65% O=25.75% S=13.46%
1
Expert's answer
2019-10-22T05:44:01-0400

To determine empirical formula,first divide the percentage of elements by their molar mass to find their relative number of moles

Relative number of moles of "C=\\frac{42.86}{12}=3.57167"

Relative number of moles of "H=\\frac{2.4}{1}=2.4"

Relative number of moles of "N=\\frac{5.88}{14}=0.42"

Relative number of moles of "Na=\\frac{9.65}{23}=0.41956\\approx0.42"

Relative number of moles of "O=\\frac{25.75}{16}=1.60"

Relative number of moles of "S=\\frac{13.46}{32}=0.42"

Then divide the relative number of moles by the smallest one.

For "C\\equiv \\frac{3.57167}{0.42}=8.5"

For "H\\equiv \\frac{2.4}{0.42}=5.71"

For "N\\equiv \\frac{0.42}{0.42}=1"

For "Na\\equiv \\frac{0.42}{0.42}=1"

For "O\\equiv \\frac{1.60}{0.42}=3.81"

For "S\\equiv \\frac{0.42}{0.42}=1"

The equivalent ratio is

"C:H:O:Na:S:N\\equiv"

"8.5:5.7:3.8:1:1:1\\equiv"

"1.9(4.5:3:2):1:1:1\\approx"

"2(4.5:3:2):1:1:1\\equiv"

"9:6:4:1:1:1"

Hence the empirical formula will be "C_9H_6O_4NaSN"


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