Question

I need help with this assignment

Answer 1

The answers for the following sums is given below.

1.$Pd_{2}H_{2}$

2.$C_{2} H_{6}$

3.$C_{2} H_{2} O_{2} Cl_{2}$

4.$C_{3}Cl_{3} N_{3}$

5.$Tl_{2 } C_{4} H_{4}O_{6}$

6.$C_{8}H_{8}$

7. $N_{2}O_{5}$

8.$P_{4}O_{6}$

9.$C_{4}H_{8} O_{2}$

Explanation:

1.Given:

        Molar mass=216.8g

Molecular formula=Pd$H_{2}$

       we know;

Molecular formula=n(Empirical formula)

molecular weight of palladium(Pd)=106.4u

molecular weight of hydrogen(H)=1u

Molar mass of Pd$H_{2}$:

Pd=106.4×1=106.4u

H=1×2=2

molar mass of Pd$H_{2}$=106.4+2=108.4

n=$\frac{216.8}{106.4}$

n=2

Molecular formula=2(Pd$H_{2}$)

Molecular formula=$Pd_{2}H_{2}$

Therefore the molecular formula of the compound is $Pd_{2}H_{2}$

2. Given:

        Molar mass=30.0g

Molecular formula=$CH_{3}$

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

Molar mass of $CH_{3}$:

C=12.01 × 1 = 12.01u

H=1 × 3 = 3u

molar mass of $CH_{3}$=12.01 + 3 =15.01u

n=$\frac{30.0}{15.01}$

n=2

Molecular formula=2($CH_{3}$)

Molecular formula=$C_{2} H_{6}$

Therefore the molecular formula of the compound is $C_{2} H_{6}$

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=$\frac{129}{64.5}$

n=2

Molecular formula=2(CHOCl)

Molecular formula=$C_{2} H_{2} O_{2} Cl_{2}$

Therefore the molecular formula of the compound is $C_{2} H_{2} O_{2} Cl_{2}$

5. Given:

        Molar mass=577g

Molecular formula=$TlC_{2} H_{2}O_{3}$

       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 $TlC_{2} H_{2}O_{3}$ :

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 $TlC_{2} H_{2}O_{3}$=204.3+24.02+1+48.00=278.32u

n=$\frac{577}{278.32}$

n=2

Molecular formula=2 ($TlC_{2} H_{2}O_{3}$)

Molecular formula=$Tl_{2 } C_{4} H_{4}O_{6}$

Therefore the molecular formula of the compound is $Tl_{2 } C_{4} H_{4}O_{6}$

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=$\frac{184.5}{61.5}$

n=3

Molecular formula=3 (CClN)

Molecular formula=$C_{3}Cl_{3} N_{3}$

Therefore the molecular formula of the compound is $C_{3}Cl_{3} N_{3}$

6. For the table refer the attached file.

Simplest ratio of elements:

Carbon=8

Hydrogen=8

Empirical formula=$C_{8}H_{8}$

Molecular formula =$C_{8}H_{8}$

Molar mass of $C_{8}H_{8}$:

molecular weight of carbon=12.04u

molecular weight of hydrogen=1u

C=8×12.01=96.08u

H=1×8=8u

molar mass of $C_{8}H_{8}$=96.08+8=104.08u

n=104.08÷78.0

n=1

Molecular formula = n(Empirical formula)

Molecular formula = 1($C_{8}H_{8}$)

Molecular formula =$C_{8}H_{8}$

Therefore the molecular formula of a compound is $C_{8}H_{8}$

7. Given:

mass of oxide of nitrogen=108g

mass of nitrogen=4.02g

mass of oxygen=11.48g

moles of nitrogen=$\frac{4.04}{14.01}$ = 0.289 moles

moles of oxygen=$\frac{11.46}{15.999}$ =0.716 moles

We divide through by the lowest molar quantity to give an empirical formula  of $N_{2} O{5}$.

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 $N_{2}O_{5}$.

8.For the table refer the attached file.

Simplest ratio of elements:

Phosphorus=2

Oxygen=3

We know;

Empirical formula=$P_{2} O_{3}$

molecular formula= 2(Empirical formula)

Molecular formula =2($P_{2} O_{3}$)

Molecular formula =$P_{4}O_{6}$

Therefore the molecular formula of the compound is $P_{4}O_{6}$

9. For the table refer the attached file.

Simplest ratio of elements:

Carbon=2

Hydrogen=9

Oxygen=2

We know;

Empirical formula =$C_{2} H_{4} O$

Molecular formula = 2(Empirical formula)

Molecular formula =2($C_{2} H_{4} O$)

Molecular formula =$C_{4}H_{8} O_{2}$

Therefore the molecular formula of the compound is $C_{4}H_{8} O_{2}$