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Alja [10]
3 years ago
8

An unknown compound contains only C , H , and O . Combustion of 4.20 g of this compound produced 10.3 g CO 2 and 4.20 g H 2 O .

What is the empirical formula of the unknown compound? Insert subscripts as needed.
Chemistry
1 answer:
vodomira [7]3 years ago
5 0

Answer:

C4H8O

Explanation:

We can get the answer through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 10.3/44 = 0.2341

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above

The mass of carbon in the compound is simply the number of moles multiplied by the atomic mass unit. The atomic mass unit of carbon is 12. The mass of carbon in the compound is thus 12 * 0.2341= 2.8091

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 4.2/18 = 0.2333

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.2333 = 0.4667 moles

The mass of hydrogen is thus 0.4667* 1 = 0.4667

The mass of oxygen equals the mass of the compound minus that of hydrogen and that of carbon.

= 4.2 - 0.4667 - 2.8091 = 0.9242

The number of moles of oxygen is the mass of oxygen divided by its atomic mass unit.

That equals 0.9242/16 = 0.0577625 moles

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that of oxygen

H = 0.4667/0.0577625 = 8

O = 0.0577625/0.0577625 = 1

C = 0.2341/0.0577625 = 4

The empirical formula is thus C4H8O

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Answer: The partial pressure of neon is 173 mmHg

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

p_{total}=p_A+p_B+p_C

Given : p_{total} =total pressure of gases = 490 mm Hg

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