# 1. The percent mass compositio

1. The percent mass composition of an unknown substance is41.37% C, 10.41% H, and 48.22% N. If we have 100 g of this chemicalcompound, what is the mass of nitrogen in this sample?

2.The percent mass composition of an unknown substance is 41.37%C, 10.41% H, and 48.22% N. If we have 100 g of this chemicalcompound, how man moles of carbon do we have in the sample?

3. The percent mass composition of an unknown substance is41.37% C, 10.41% H, and 48.22% N. If we have 100 g of this chemicalcompound, what is the ratio of moles of hydrogen to moles ofnitrogen?

4. The percent mass composition of an unknown substance is41.37% C, 10.41% H, and 48.22% N. What is the empirical formula ofthis compound?

5. The percent mass composition of an unknown substance is41.37% C, 10.41% H, and 48.22% N. If the molar mass of thiscompound is 58.07 g/mol, what is the ratio of the total number ofatoms in the molecular formula to the total number of atoms in theempirical formula?

Percent composition is the percent by mass of each element in acompound. It is given by the formula :

%Mass = Mass of an element present in 1 mole of thecompound/Total molecular mass of the compound * 100 %

Composition of the given compound :

C = 41.37%

H = 10.41%

N = 48.22%

This means that for 1g of the compound,

Mass of C = 0.4137 g and so on

1. Therefore, for 100g of the compound :

Mass of C = 41.37g

Mass of H = 10.41g

Mass of N = 48.22g

Mass of N present in the 100g sample= 48.22g

2. Molar mass of C = 12g/mol

Mass of C present in 100g sample =41.37g

No. of moles of C = Mass of C present/ Molar mass of C =41.37/12 = 3.45 moles

No. of moles of C present in the 100g sample = 3.45 moles

3.Molar mass of H = 1g/mol

Molar mass of N = 14g/mol

No. of moles of H present = 10.41/1 = 10.41

No. of moles of N present = 48.22/14 = 3.44

Ratio of H to N = 10.41/3.44 = 3:1

4.Empirical formula represents the simplest set of whole numbersexpressing the relative numbers of atoms in the compound, whichalso represents the relative numbers of moles of atoms.

So, by calculating the ratio of number of moles of each elementin the compound, we can easily determine the empirical formula ofthe compound. The ratio is rounded off to the nearest whole numberratio.

For 100g of the sample,

No. of moles of C = 3.45

No. of moles of H = 10.41

No. of moles of N = 3.44

C:H:N = 3.45:10.41:3.44 = 1:3:1

Therefore, the empirical formula is CH3N

5.The molecular formula weight is a whole number multiple of theempirical formula weight for a given compound.

Molecular formula weight/ Empirical formula weight= n(say)

Then if X represents the empirical formula of the compound, itsmolecular formula will be (X)n

Given, molar mass of the compound = 58.07g/mol

Empirical formula weight = 12 + 3*1 + 14 = 29g

n = 58.07/29 = 2

Therefore, molecular formula is (CH3N)2 =C2H6N2

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