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27 August, 15:02

What is the empirical formula of a compound that contains 6.10 g of hydrogen and 28 g of nitrogen?

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  1. 27 August, 16:01
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    NH₃

    Explanation:

    mass H = 6.10 grams

    mass N = 28.00 grams

    mass cpd = (6.10 + 28.00) grams = 34.10 grams

    %H/100wt = (6.10/34.10) 100% = 17.9% w/w

    %N/100wt = (28.00/34.1) 100% = 82.1% w/w

    %/100wt = > grams/100wt = > moles = > ratio = > reduce = > emp ratio

    %H/100wt = 17.9% w/w = > 17.9g = > (17.9/1) moles = 17.9 moles H

    %N/100wt = 82.1% w/w = > 82.1g = > (82.1/14) moles = 5.9 moles N

    Ratio N:H = > 17.9 : 5.9

    Reduce mole ratio (divide by smaller mole value) = > 17.9/5.9 : 5.9/5.9

    => 3HY:1H empirical ratio = > empirical formula NH₃ (ammonia)
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