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Anika [276]
3 years ago
5

What is the empirical formula of a compound that breaks down into 4.12g of N and 0.88g of H? N 5 H NH 3 NH 4 N 4 H

Chemistry
1 answer:
Marysya12 [62]3 years ago
4 0

<u>The </u><u>emperical formula </u><u>can be </u><u>calculated </u><u>as follows:</u>


<u>mass :</u>   4.12g of N            and        0.88g of H

<u>moles:</u>   4.12g/14.007       and        0.88g/1.008

               

<u>moles:</u>   0.294                   and        0.873


<u>moles/smallest number</u>:   (gives the ratio of atoms)


             0.294/0.294                and           0.873/0.294

                   1                               and             2.969

                    1                              and              3



The simplest ratio of N : H is 1 : 3

<u>Hence the </u><u>emperical formula </u><u>is </u><u>NH3</u>


<u>Explanation:</u>

 1.   The emperical formula can be found by getting  the <u>simplest ratio of atoms in a molecule.</u>

2. To get the simplest ratio of atoms we find out the <u>number of moles of each atoms in it.</u>

3. From the number of moles of atoms we can find the simplest ratio of each atom by dividing the <u>number of moles by the smallest number.</u>


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