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LUCKY_DIMON [66]
3 years ago
8

An ionic compound is composed of the following elements: Nitrogen – 31.57% Hydrogen – 9.10% Phosphorus – 23.27% Oxygen – 36.06%

What is the empirical formula for this compound?
Chemistry
2 answers:
malfutka [58]3 years ago
7 0

Answer:

N3H12PO3

Explanation:

Step 1:

Data obtained from the question.

Nitrogen (N) = 31.57%

Hydrogen (H) = 9.10%

Phosphorus (P) = 23.27%

Oxygen (O) = 36.06%

Step 2:

Divide by their molar masses.

This is illustrated below:

N = 31.57/14 = 2.255

H = 9.10/ 1 = 9.1

P = 23.27/31 = 0.751

O = 36.06/16 = 2.254

Step 3:

Divide by the smallest number. This is illustrated below:

N = 2.255/0.751 = 3

H = 9.1 /0.751 = 12

P = 0.751 /0.751 = 1

O = 2.254/0.751 = 3

The empirical formula is N3H12PO3

djverab [1.8K]3 years ago
3 0

Answer:

The empirical formula is: <u>N3H12PO3</u>

Explanation:

To calculate the empirical formula, the values ​​of each element must first be taken in 100gr of the compound:

  • 31.57% Nitrogen = <u>31.57gr N </u>
  • 9.10% Hydrogen = <u>9.10 gr H </u>
  • 23.27% Phosphorus = <u>23.27gr P </u>
  • 36.06% Oxygen = <u>36.06gr O </u>

Then the grams of each element are divided by their atomic weight:

  • 31.57grN / 14U =<u> 2,255 N </u>
  • 9.10grH / 1.001U = <u>9.091 H </u>
  • 23.27grP / 31U = <u>0.75 P </u>
  • 36.06grO / 16U = <u>2.253 O </u>

Then each data is divided by the smallest number that is that of Phosphorus:

  • 2,255 N / 0.75 = <u>3N </u>
  • 9,091 H / 0.75 = <u>12H </u>
  • 0.75 P / 0.75 = <u>1 P </u>
  • 2,253 O / 0.75 = <u>3 O </u>

And the data we use is put into a formula of the compound:

  • NHPO

And its values ​​are assigned:

  • <u>N3H12PO3</u>
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72.0 mL of steam is formed.

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OR

Just transform the chemical equation by dividing the whole equation by 4 so that the coefficient of NH_{3} become one like this

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<em>1.  </em><em>Assume ammonia to be completely exhausted</em>

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2.  <em>Assume oxygen to be completely exhausted</em>

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Now we know that during complete reaction 48 mL of ammonia and 60 mL of oxygen is used which will form 60 \times \frac{4}{5} (= 48) mL of nitic oxide gas and 60 \times \frac{6}{5} (= 72) mL of steam.

Therefore <em>72 mL of steam </em>is formed.

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