Concept explainers
(a)
Interpretation:
The empirical formula the compound with percentage composition of 63.6% N &36.4% O has to be given.
Concept Introduction:
Empirical Formula:
The empirical formula of a compound is the simplest whole number ratio of each type of atom in a compound. It can be the same as the compound’s molecular formula but not always. An empirical formula can be calculated from information about the mass of each element in a compound or from the percentage composition.
The steps for determining the empirical formula of a compound as follows:
- Obtain the mass of each element present in grams.
- Determine the number of moles of each atom present.
- Divide the number of moles of each element by the smallest number of moles.
- Convert the numbers to whole numbers. The set of whole numbers are the subscripts in the empirical formula.
(a)
Answer to Problem 35PE
The empirical formula of the compound is N2O.
Explanation of Solution
Given,
The percentage composition of nitrogen is 63.6%.
The percentage composition of oxygen is 36.4%.
The
The atomic mass of oxygen is 16 g/mol.
Assuming that 100 g of material, the percent of each element equals the grams of each element.
The grams of each element has to be converted to moles as,
The moles of nitrogen =(63.6 g)×1 mol14.01 g= 4.54mol
The moles of oxygen =(36.4 g)×1 mol16 g=2.28 mol
The empirical formula can be calculated as,
The number of moles can be converted to moles to whole numbers by dividing by the small number.
N=4.54 mol2.28 mol=2
O=2.28 mol2.28 mol=1
The empirical formula of the compound is N2O.
(b)
Interpretation:
The empirical formula the compound with percentage composition of 46.71% N & 53.3% O has to be given.
Concept Introduction:
Refer to part (a).
(b)
Answer to Problem 35PE
The empirical formula of the compound is NO.
Explanation of Solution
Given,
The percentage composition of nitrogen is 46.71%.
The percentage composition of oxygen is 53.3%.
The atomic mass of nitrogen is 14.01 g/mol.
The atomic mass of oxygen is 16 g/mol.
Assuming that 100 g of material, the percent of each element equals the grams of each element.
The grams of each element has to be converted to moles as,
The moles of nitrogen =(46.7 g)×1 mol14.01 g= 3.33mol
The moles of oxygen =(53.3 g)×1 mol16 g=3.33 mol
The empirical formula can be calculated as,
The number of moles can be converted to moles to whole numbers by dividing by the small number.
N=3.33 mol3.33 mol=1
O=3.33 mol3.33 mol=1
The empirical formula of the compound is NO.
(c)
Interpretation:
The empirical formula the compound with percentage composition of 25.9% N &74.1% O has to be given.
Concept Introduction:
Refer to part (a).
(c)
Answer to Problem 35PE
The empirical formula is N2O5.
Explanation of Solution
Given,
The percentage composition of nitrogen is 25.9%.
The percentage composition of oxygen is 74.1%.
The atomic mass of nitrogen is 14.01 g/mol.
The atomic mass of oxygen is 16 g/mol.
Assuming that 100 g of material, the percent of each element equals the grams of each element.
The grams of each element has to be converted to moles as,
The moles of nitrogen =(25.9 g)×1 mol14.01 g= 1.85mol
The moles of oxygen =(74.1 g)×1 mol16 g=4.63 mol
The empirical formula can be calculated as,
The number of moles can be converted to moles to whole numbers by dividing by the small number.
N=1.85 mol1.85 mol=1
O=4.63 mol1.85 mol=2.5
Since the value is not a whole number multiply each by two. The empirical formula is N2O5.
(d)
Interpretation:
The empirical formula the compound with percentage composition of 43.4% Na, 11.3%C &45.3% O has to be given.
Concept Introduction:
Refer to part (a).
(d)
Answer to Problem 35PE
The empirical formula is Na2CO3.
Explanation of Solution
Given,
The percentage composition of sodium is 43.4%.
The percentage composition of oxygen is 45.3%.
The percentage composition of carbon is 11.3%.
The atomic mass of sodium is 22.99 g/mol.
The atomic mass of oxygen is 16 g/mol.
The atomic mass of carbon is 12.01 g/mol.
Assuming that 100 g of material, the percent of each element equals the grams of each element.
The grams of each element has to be converted to moles as,
The moles of sodium =(43.4 g)×1 mol22.99 g= 1.89mol
The moles of oxygen =(45.3 g)×1 mol16 g=2.83 mol
The moles of carbon =(11.3 g)×1 mol12.01 g= 0.941mol
The empirical formula can be calculated as,
The number of moles can be converted to moles to whole numbers by dividing by the small number.
Na=1.89 mol0.941 mol=2.0
C=0.941 mol0.941 mol=1.0
O=2.83 mol0.941 mol=3.0
The empirical formula is Na2CO3.
(e)
Interpretation:
The empirical formula the compound with percentage composition of 18.8% Na, 29.0%Cl &52.3% O has to be given.
Concept Introduction:
Refer to part (a).
(e)
Answer to Problem 35PE
The empirical formula is NaClO4.
Explanation of Solution
Given,
The percentage composition of sodium is 18.8%.
The percentage composition of oxygen is 53.3%.
The percentage composition of chlorine is 29%.
The atomic mass of sodium is 22.99 g/mol.
The atomic mass of oxygen is 16 g/mol.
The atomic mass of chlorine is 35.45 g/mol.
Assuming that 100 g of material, the percent of each element equals the grams of each element.
The grams of each element has to be converted to moles as,
The moles of sodium =(18.8 g)×1 mol22.99 g= 0.818mol
The moles of oxygen =(52.3 g)×1 mol16 g=3.27 mol
The moles of chlorine =(29 g)×1 mol35.45 g= 3.27mol
The empirical formula can be calculated as,
The number of moles can be converted to moles to whole numbers by dividing by the small number.
Na=0.818 mol0.818 mol=1.0
Cl=0.818 mol0.818 mol=1.0
O=3.27 mol0.818 mol=4.0
The empirical formula is NaClO4.
(f)
Interpretation:
The empirical formula the compound with percentage composition of 72.02% Mn &27.98 % O has to be given.
Concept Introduction:
Refer to part (a).
(f)
Answer to Problem 35PE
The empirical formula is Mn3O4.
Explanation of Solution
Given,
The percentage composition of manganese is 72.02%.
The percentage composition of oxygen is 27.98%.
The atomic mass of manganese is 54.94 g/mol.
The atomic mass of oxygen is 16 g/mol.
Assuming that 100 g of material, the percent of each element equals the grams of each element.
The grams of each element has to be converted to moles as,
The moles of manganese =(72.02 g)×1 mol54.94 g= 1.311 mol
The moles of oxygen =(27.98 g)×1 mol16 g=1.749 mol
The empirical formula can be calculated as,
The number of moles can be converted to moles to whole numbers by dividing by the small number.
Mn=1.311 mol1.311 mol=1.0
O=1.749 mol1.311 mol=1.334
Since the value is not a whole number multiply each by three. The empirical formula is Mn3O4.
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Chapter 7 Solutions
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