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?
2 answers:
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
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:
And its values are assigned:
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