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aliina [53]
2 years ago
9

What is the empirical formula of a compound with the following percent composition: 18% C, 2.5% H, 63.5% I, and 16.0% O?

Chemistry
1 answer:
borishaifa [10]2 years ago
7 0

The empirical formula is C₃H₅ IO₂

<u>Explanation:</u>

Assume 100 g of the compound is present. This changes the percents to grams:  

C = 18 g, H = 2.5 g , I = 63. 5 g , O = 16 g

Convert the masses to moles:

C = 18/ 12 = 1.5

H = 2.5/1 = 2.5

I = 63.5/ 127 =0.5

O = 16/16 = 1

Divide by the lowest, seeking the smallest whole-number ratio:

C = 1.5 / 0.5 = 3

H = 2.5/0.5=5

I = 0.5/0.5= 1

O = 1/0.5= 2

Now we can write the Empirical formula as,

C₃H₅ IO₂

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Answer:

Explanation:

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As per your description:

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If the maximum possible amount of NH₃ is formed during the reaction, you assume that the reaction goes to completion.

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Since, the squares show that there are 3 molecules of each reactant, the 3 molecules of hydrogen gas will be able to react with 1 molecule of nitrogen gas. When that happens, all the hydrogen gas is consumend and yet two molecules of nitrogen gas will remain unreacted. Hence, the nitrogen gas is the leftover reactant.

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