Combustion analysis of a compound containing n: while the composition of carbon and hydrogen can be determined directly from combustion analysis, elements like n and s have to be determined by separate methods. use information provided for next 5 questions: a compound contains only carbon, hydrogen, nitrogen, and oxygen. combustion of a 2.18 g sample burns in excess oxygen yields 3.94 g of co2 and 1.89 g of h2o. a separate experiment shows that a 1.23 g sample contains 0.235 g of n. calculate the moles of c in the sample.

0.1 moles

Explanation:

Okay, the explanation here is pretty long. Let’s go!

Firstly, we know that the compound contains only carbon, hydrogen, oxygen and nitrogen. So the molecular formula would look like this CxHyOzNp

Where x , y, z and p are number of atoms of each element present respectively. According to the law of constant composition, the ratio of the number of atoms are fixed irrespective of the source or method of preparation. From this, we now know that in both samples, we have the same number of atoms. What is proper to do is to calculate the numbers, we do that as follows.

Firstly, we will need to calculate these numbers using the masses given in the first sample. Over the calculations, we should note that the formula we would be using is the relation: mass = number of moles * atomic mass or molar mass. Rearranging the equation gives different variations of the formula.

Now let’s do some mathematics.

There is 3.94g of carbon iv oxide, we can calculate the number of moles of it present which eventually would yield the number of moles of carbon present.

The molar mass of carbon iv oxide is 44g/mol.

The number of moles of carbon iv oxide present is thus 3.94/44 = 0.0895 moles

Since there is just 1 atom of carbon present in carbon iv oxide, this means the number of moles of carbon present is also 0.0895 moles

The mass of carbon present is the number of moles of it present multiplied by the atomic mass unit of carbon which is 12. This mass is 0.0895 * 12 = 1.0745g

Next, we calculate the number of moles of hydrogen and consequently its mass present.

To get this, we can access it from the number of moles of water present.

We get this by dividing the mass of water present by the molar mass of water. This is equals 1.89/18 = 0.105 moles. Now we know that 1 mole of water has 2 atoms of hydrogen, hence 1 mole of water will yield 2 moles of hydrogen. The number of moles of hydrogen present is thus 0.105 * 2 = 0.21 moles.

The mass of hydrogen thus present is 0.21 * 1 = 0.21g

Now, we know the mass of hydrogen present and the mass of carbon present. But, we do not know the mass of oxygen and nitrogen present.

To get this, we subtract the mass of hydrogen and carbon present from the mass of the total= 2.18 - 1.0745 - 0.21 = 0.8955g

Now we know the mass of oxygen and nitrogen combined. We can access their number of moles using their respective atomic masses. The total number of moles present is equal 0.8955/30 = 0.02985 moles

Wondering where 30 came from? The atomic mass of nitrogen and oxygen are 14 and 16 respectively. We now get the number of moles of both present. This is equal to the atomic mass divided by the total mass multiplied by the number of moles.

For oxygen = 16/30 * 0.02985 = 0.01592

For nitrogen = 14/30 * 0.02985 = 0.01393

From this we can try and get an empirical formula for the compound. This helps us to know the ratios of the number of atoms. To get this , we divide the number of moles by the smallest number of moles. The smallest number of moles is unarguably that of nitrogen. The empirical formula is calculated as follows:

C = 0.0895/0.01393 = 6

H = 0.21/0.01393 = 15

O = 0.01592/0.01393 = 1

N = 0.01393/0.01393 = 1

Thus, the empirical formula looks like this :

C6H15NO

Now, we can move to the second sample.

We know that the sample contains 0.235g of nitrogen. We first need to get the number of moles of nitrogen present in the sample. To get this, we simply divide this mass by the atomic mass. That is: 0.235/14 = 0.0168 moles

Now since the question asks us to get the number of moles of carbon and we know that in any elemental analysis, the ratio of carbon to nitrogen is 6 to 1, we simply multiply the number of moles of nitrogen by 6.

Hence, this is 0.0168 * 6 = 0.1 moles

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