The answers for the following sums is given below.
1.[tex]Pd_{2}H_{2}[/tex]
2.[tex]C_{2} H_{6}[/tex]
3.[tex]C_{2} H_{2} O_{2} Cl_{2}[/tex]
4.[tex]C_{3}Cl_{3} N_{3}[/tex]
5.[tex]Tl_{2 } C_{4} H_{4}O_{6}[/tex]
6.[tex]C_{8}H_{8}[/tex]
7. [tex]N_{2}O_{5}[/tex]
8.[tex]P_{4}O_{6}[/tex]
9.[tex]C_{4}H_{8} O_{2}[/tex]
Explanation:
1.Given:
Molar mass=216.8g
Molecular formula=Pd[tex]H_{2}[/tex]
we know;
Molecular formula=n(Empirical formula)
molecular weight of palladium(Pd)=106.4u
molecular weight of hydrogen(H)=1u
Molar mass of Pd[tex]H_{2}[/tex]:
Pd=106.4×1=106.4u
H=1×2=2
molar mass of Pd[tex]H_{2}[/tex]=106.4+2=108.4
n=[tex]\frac{216.8}{106.4}[/tex]
n=2
Molecular formula=2(Pd[tex]H_{2}[/tex])
Molecular formula=[tex]Pd_{2}H_{2}[/tex]
Therefore the molecular formula of the compound is [tex]Pd_{2}H_{2}[/tex]
2. Given:
Molar mass=30.0g
Molecular formula=[tex]CH_{3}[/tex]
we know;
Molecular formula = n (Empirical formula)
molecular weight of Carbon(C)=12.01u
molecular weight of hydrogen(H)=1u
Molar mass of [tex]CH_{3}[/tex]:
C=12.01 × 1 = 12.01u
H=1 × 3 = 3u
molar mass of [tex]CH_{3}[/tex]=12.01 + 3 =15.01u
n=[tex]\frac{30.0}{15.01}[/tex]
n=2
Molecular formula=2([tex]CH_{3}[/tex])
Molecular formula=[tex]C_{2} H_{6}[/tex]
Therefore the molecular formula of the compound is [tex]C_{2} H_{6}[/tex]
3. Given:
Molar mass=129g
Molecular formula=CHOCl
we know;
Molecular formula = n (Empirical formula)
molecular weight of Carbon(C)=12.01u
molecular weight of hydrogen(H)=1u
molecular weight of oxygen(O)=16.00u
molecular weight of chlorie(Cl)=35.5u
Molar mass of CHOCl:
C=12.01 × 1 = 12.01u
H=1 × 1 = 1u
O=16.00×1=16.00u
Cl=35.5×1=35.5u
molar mass of CHOCl=12.01+1+16.00+35.5=64.5u
n=[tex]\frac{129}{64.5}[/tex]
n=2
Molecular formula=2(CHOCl)
Molecular formula=[tex]C_{2} H_{2} O_{2} Cl_{2}[/tex]
Therefore the molecular formula of the compound is [tex]C_{2} H_{2} O_{2} Cl_{2}[/tex]
5. Given:
Molar mass=577g
Molecular formula=[tex]TlC_{2} H_{2}O_{3}[/tex]
we know;
Molecular formula = n (Empirical formula)
molecular weight of Carbon(C)=12.01u
molecular weight of Thallium(Tl)=204.3u
molecular weight of hydrogen(H)=1u
molecular weight of oxygen(O)=16.00u
Molar mass of [tex]TlC_{2} H_{2}O_{3}[/tex] :
C=12.01 × 2= 24.02u
Tl=204.3×1=204.3u
H=1×2=2u
O=16.00×3=48.00
molar mass of [tex]TlC_{2} H_{2}O_{3}[/tex]=204.3+24.02+1+48.00=278.32u
n=[tex]\frac{577}{278.32}[/tex]
n=2
Molecular formula=2 ([tex]TlC_{2} H_{2}O_{3}[/tex])
Molecular formula=[tex]Tl_{2 } C_{4} H_{4}O_{6}[/tex]
Therefore the molecular formula of the compound is [tex]Tl_{2 } C_{4} H_{4}O_{6}[/tex]
4. Molar mass=184.5g
Molecular formula=CClN
we know;
Molecular formula = n (Empirical formula)
molecular weight of Carbon(C)=12.01u
molecular weight of Nitrogen(N)=14u
molecular weight of chlorine(Cl)=35.5u
Molar mass of CClN:
C=12.01 × 1 = 12.01u
N=1×14=14U
Cl=35.5×1=35.5u
molar mass of CClN=12.01+14+35.5=61.5u
n=[tex]\frac{184.5}{61.5}[/tex]
n=3
Molecular formula=3 (CClN)
Molecular formula=[tex]C_{3}Cl_{3} N_{3}[/tex]
Therefore the molecular formula of the compound is [tex]C_{3}Cl_{3} N_{3}[/tex]
6. For the table refer the attached file.
Simplest ratio of elements:
Carbon=8
Hydrogen=8
Empirical formula=[tex]C_{8}H_{8}[/tex]
Molecular formula =[tex]C_{8}H_{8}[/tex]
Molar mass of [tex]C_{8}H_{8}[/tex]:
molecular weight of carbon=12.04u
molecular weight of hydrogen=1u
C=8×12.01=96.08u
H=1×8=8u
molar mass of [tex]C_{8}H_{8}[/tex]=96.08+8=104.08u
n=104.08÷78.0
n=1
Molecular formula = n(Empirical formula)
Molecular formula = 1([tex]C_{8}H_{8}[/tex])
Molecular formula =[tex]C_{8}H_{8}[/tex]
Therefore the molecular formula of a compound is [tex]C_{8}H_{8}[/tex]
7. Given:
mass of oxide of nitrogen=108g
mass of nitrogen=4.02g
mass of oxygen=11.48g
moles of nitrogen=[tex]\frac{4.04}{14.01}[/tex] = 0.289 moles
moles of oxygen=[tex]\frac{11.46}{15.999}[/tex] =0.716 moles
We divide through by the lowest molar quantity to give an empirical formula of [tex]N_{2} O{5}[/tex].
Now the molecular formula is multiple of the empirical formula.
So,
108 = n × (2×14.01 + 5×15.999)
Clearly,n=1, and the molecular formula is [tex]N_{2}O_{5}[/tex].
8.For the table refer the attached file.
Simplest ratio of elements:
Phosphorus=2
Oxygen=3
We know;
Empirical formula=[tex]P_{2} O_{3}[/tex]
molecular formula= 2(Empirical formula)
Molecular formula =2([tex]P_{2} O_{3}[/tex])
Molecular formula =[tex]P_{4}O_{6}[/tex]
Therefore the molecular formula of the compound is [tex]P_{4}O_{6}[/tex]
9. For the table refer the attached file.
Simplest ratio of elements:
Carbon=2
Hydrogen=9
Oxygen=2
We know;
Empirical formula =[tex]C_{2} H_{4} O[/tex]
Molecular formula = 2(Empirical formula)
Molecular formula =2([tex]C_{2} H_{4} O[/tex])
Molecular formula =[tex]C_{4}H_{8} O_{2}[/tex]
Therefore the molecular formula of the compound is [tex]C_{4}H_{8} O_{2}[/tex]
