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alukav5142 [94]
3 years ago
5

An unknown compound was found to have a percent composition as follows: 47.0%, 14.5 carbon, and 38.5 oxygen. What is its empiric

al formula? If the true molar mass of the compound is 166.22 g/mol, what is its molecular formula?
Chemistry
1 answer:
shusha [124]3 years ago
6 0

 KCO₂

 K₂C₂O₄

Explanation:

Given parameters:

Percent composition:

             K = 47%

             C = 14.5%

             O = 38.5%

Molar mass of compound = 166.22g/mol

Unknown:

Empirical formula of compound = ?

Molecular formula of compound = ?

Solution:

The empirical formula of a compound is its simplest formula. Here is how to solve for it:

 

                                K                                C                           O

Percent

composition           47                               14.5                       38.5

Molar mass            39                                  12                           16

Number

of moles               47/39                         14.5/12                     38.5/16

moles                    1.205                          1.208                          2.4

Dividing

by smallest      1.205/1.205               1.208/1.205                      2.4/1.205

                                1                                  1                                        2

Empirical formula               KCO₂

Molecular formula

  This is the actual combination of the atoms:

          Molecular formula =   ( empirical formula of KCO₂)ₙ

 Molar mass of empirical formula = 39 + 12 + 2(16) = 83g/mol

      n factor = \frac{true molecular mass}{molar mas of empirical formula}

      n factor = \frac{166.22}{83} =  2

Molecular formula of compound = ( KCO₂)₂ = K₂C₂O₄

Learn more:

Empirical formula brainly.com/question/2790794

#learnwithBrainly

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<h3>For the calculation of molarity of solution</h3>

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