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-BARSIC- [3]
2 years ago
8

What is empirical formula if compound consists of 21.2%N, 6.1%H, 24.2% and 48.5%O?

Chemistry
1 answer:
vichka [17]2 years ago
5 0

Answer:

                      Empirical Formula  =  N₂H₈SO₄   Or    (NH₄)₂SO₄

Explanation:

Step 1: Calculate Moles of each Element;

                     Moles of N  =  %N ÷ At.Mass of N

                     Moles of N  = 21.20 ÷ 14.01

                     Moles of N  =  1.5132 mol

                     Moles of H  =  %H ÷ At.Mass of H

                     Moles of H  = 6.10 ÷ 1.01

                     Moles of H  =  6.0396 mol

                     Moles of S  =  %S ÷ At.Mass of S

                     Moles of S  = 24.20 ÷ 32.06

                     Moles of S  =  0.7548 mol

                     Moles of O  =  %O ÷ At.Mass of O

                     Moles of O  = 48.50 ÷ 16.0

                     Moles of O  =  3.0312 mol

Step 2: Find out mole ratio and simplify it;

               N                                        H                                     S

           1.5132                              6.0396                             0.7548

    1.5132/0.7548                  6.0396/0.7548                 0.7548/0.7548

              <u>2</u>                                        <u>8</u>                                       <u>1</u>

                                                    O

                                                  3.0312

                                           3.0312/0.7548

                                                     <u>4</u>

Result:

       Empirical Formula  =  N₂H₈SO₄   Or    (NH₄)₂SO₄

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