Answer:
1. Percent composition of Al = 13.423 %
2.
- Percent composition of Zn = 28.02 %
- Percent composition of Cl = 30.6 %
- Percent composition of O = 41.3 %
3. The empirical formula is C₅O₁₆
4. Molecular Formula= P₄O₆
Explanation:
Part first :
Data Given
Formula of the Molecule = Al₂ (CrO₄)₃
% of Al₂ = ?
> First of all find the atomic masses of each component in a molecule
For Al₂ (CrO₄)₃ atomic masses are given below
Al = 27 g/mol
Cr = 52 g/mol
O = 16 g/mol
> Then find the total masses of each component
2 atoms of Al = 27 g/mol x 2
= 54 g/mol
3 atoms of Cr = 52 g/mol x 3
= 156 g/mol
12 atoms of O = 16 g/mol x 12
= 192 g/mol
> find total Molar Mass of Molecule:
Molar Mass of Al₂ (CrO₄)₃ = [27x2 + 52x3 + 16x12]
Molar Mass of Al₂ (CrO₄)₃ = 402
Now to find the mass percent of Al
Formula used to find the Mass percent of a component
Percent composition of Al = mass of Al in Molecula / molar mass of Al₂(CrO₄)₃ x 100%
Put the values
Percent composition of Al = 54 (g/mol) / 402 (g/mol) x 100%
Percent composition of Al = 13.423 %
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Part 2
Data Given
Formula of the Molecule = Zn(ClO₃)₂
% Zn = ?
% Cl = ?
% O = ?
> First of all find the atomic masses of each component in a molecule
For Zn(ClO₃)₂ atomic masses are given below
Zn = 65 g/mol
Cl = 35.5 g/mol
O = 16 g/mol
> Then find the total masses of each component
1 atoms of Zn= 65 g/mol x 1
= 65 g/mol
2 atoms of Cl = 35.5 g/mol x
= 71 g/mol
6 atoms of O = 16 g/mol x 6
= 96 g/mol
> find total Molar Mass of Molecule:
Molar Mass of Zn(ClO₃)₂ = [65x1 + 35.5x2 + 16x6]
Molar Mass of Zn(ClO₃)₂ = 232g/mol
Now to find the mass percent of of each component one by one
1. Formula used to find the mass percent of Zn
Percent composition of Zn= mass of Zn in Molecular / molar mass of Zn(ClO₃)₂ x 100%
Put the values
Percent composition of Zn = 65(g/mol) / 232 (g/mol) x 100%
Percent composition of Zn = 28.02 %
-------------------
2. Formula used to find the mass percent of Cl
Percent composition of Cl = mass of Cl in Molecular / molar mass of Zn(ClO₃)₂ x 100%
Put the values
Percent composition of Cl = 71 (g/mol) / 232 (g/mol) x 100%
Percent composition of Cl = 30.6 %
---------------------
3. Formula used to find the mass percent of O
Percent composition of O = mass of O in Molecular / molar mass of Zn(ClO₃)₂ x 100%
Put the values
Percent composition of O = 96 (g/mol) / 232 (g/mol) x 100%
Percent composition of O = 41.3 %
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Part 3:
Data Given
Percentage of C = 27.3 %
Percentage of O = 72.7 %
Emperical Formula of the compound = ?
Solution:
So the compound has 27.3 % C and 72% O
First, find the mass of each of the elements in 100 g of the Compound.
C = 27.3 g
O = 72 g
Now find how many moles are there for each element in 100 g of compound
For this molar mass are required
That is
C = 12 g/mol
O = 16 g/mol
Formula Used
mole of C = mass of C / Molar mass of C
mole of C = 27.3 / 12 g/mol
mole of C = 2.275
Formula Used
mole of O = mass of O / Molar mass of O
mole of O = 72g / 16 g/mol
mole of O = 7.2
Divide each one by the smallest number of moles
C = 2.275 / 2.275
C = 1
O = 7.2 / 2.275
O = 3.2
Multiply the mole fraction to a number to get the whole number.
C = 1 x 5 = 5
O = 3.2 x 5 = 16
So, the empirical formula is C₅O₁₆
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Part 4
Data Given
Percentage of P= 56.38 %
Percentage of O = 43.62%
Molar Mass = 219.9g
Molecular Formula of the compound = ?
Solution:
First, find the mass of each of the elements in 100 g of the Compound.
Mass of P= 56.38g
Mass of O = 43.62g
Now find how many moles are there for each element in 100 g of compound
find the moles in total compounds
Formula Used
mole of P = mass of / Molar mass of P
mole of P = 56.38 g / 31 g/mol
mole of P = 1.818
Formula Used
mole of O = mass of O / Molar mass of O
mole of O = 43. 62 / 16 g/mol
mole of O = 2.7262
Now
first find the Emperical formula
Divide each one by the smallest number of moles
P = 1.818 /1.818
P= 1
for oxygen
O = 2.7262 / 1.818
O = 1.5
Multiply the mole fraction to a number to get the whole number.
P = 1 x 2 = 2
O = 1.5 x 2 = 3
So, the empirical formula is P₂O₃
Now
Find molar mass of the empirical formula P₂O₃
2 (31) + 3 (16) = 62 + 48 = 110
Now find that how many empirical units are in a molecular unit.
(219.9 g/mol) / ( 110 g/mol) = empirical units per molecular unit
empirical units per molecular unit = 1.999 =2
A here we get two empirical units in a molecular unit,
So the molecular formula is:
2 (P₂O₃) = P₄O₆
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