Combustion of a compound of formula cxhyoz yields 11.53 g h2o and 37.56 g co2 when 18.37 g of the compound are burned. if the molar mass of the compound is between 50 and 100 g/mol, what is the molecular formula of the compound? a) c5h8o3 b) c4h6o2 c) c4h3o4 d) c7h5o2

B

Explanation:

The statement that the molar mass is between 50 and 100 eliminates some choices as their molar masses are greater than 100. Option B is the only probable answer as regards the information given. 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 37.56/44 = 0.8536

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 * 0.8536 = 10.24g

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 11.53/18 = 0.641 moles

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.641 = 1.282 moles

The mass of hydrogen is thus 1.282 * 1 = 1.282g

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

= 18.37 - 10.24 - 1.282 = 6.848g

Now, we can calculate the number of moles of oxygen. We simply divide the mass here by the atomic mass unit of oxygen. The atomic mass unit of oxygen is 16g/mol

The number of moles of oxygen is thus 6.848/16 = 0.428 moles

Now we proceed to calculate the empirical formula of the compound. This can be done by dividing the respective number of moles by the smallest number of moles.

For Hydrogen = 1.282/0.428 = 3

For carbon = 0.8536/0.428 = 2

For oxygen = 0.428/0.428 = 1

The empirical formula is thus C2H3O

Now we know that the molecular formula is just a multiple of the empirical formula. This means the multiplying factor can be two, one or three.

One will not be possible because we will have the total mass less than 50.

Three will not be possible as the mass would be greater than 100

The only probable answer because of the range of values of the molar mass is C4H6O2, this in fact give a mass of 96g/mol which is still within the answer range.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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