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Fittoniya [83]
2 years ago
10

According to the law of constant composition, how many grams of oxygen does this isolated sample contain?

Chemistry
1 answer:
KatRina [158]2 years ago
8 0

The Mass of oxygen in isolated sample is 8.6 g

<h3>What is the Law of Constant composition?</h3>

The law of constant composition states that pure samples of the same compound contain the same element in the same ratio by mass irrespective of the source from which the compound is obtained.

Considering the given ascorbic acid samples:

Laboratory sample contains 1.50 gg of carbon and 2.00 gg of oxygen

mass ratio of oxygen to carbon is 2 : 1.5

Isolated sample will contain 2/1.5 * 6.45 g of oxygen.

Mass of oxygen in isolated sample = 8.6 g

In conclusion, the mass of oxygen is determined from the mass ratio of oxygen and carbon in the compound.

Learn more about the Law of Constant composition at: brainly.com/question/1557481

#SPJ1

Note that the complete question is given below:

A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains 1.50 g of carbon and 2.00 g of oxygen. Another sample of ascorbic acid isolated from citrus fruits contains 6.45 gg of carbon. According to the law of constant composition, how many grams of oxygen does this isolated sample contain?

Express the answer in grams to three significant figures.

8.47 g

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If a reaction is not spontaneous, what is true of the reaction?
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200.00 grams of an organic compound is known to contain 83.884 grams of carbon, 10.486
Dmitriy789 [7]

Explanation:

Mass of the organic compound = 200g

Mass of carbon = 83.884g

Mass of hydrogen = 10.486g

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The mass of nitrogen = mass of organic compound - (mass of carbon + mass of hydrogen + mass of oxygen)

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Mass of nitrogen = 86.99g

The empirical formula of a compound is its simplest formula.

It is derived as shown below;

                        C                   H                O                  N

Mass          83.884             10.486        18.64            86.99

molar

mass                12                    1                  16                14

Moles       83.884/12         10.486/1       18.64/16        86.99/14

                   

                    6.99                 10.49              1.17                6.21

Divide

by

lowest      6.99/1.17         10.49/1.17           1.17/1.17            6.21/1.17

                       6                    9                         1                       5

Empirical formula  C₆H₉ON₅

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