a compound on analysis gave the following percentage composition by ma

1.	a compound on analysis gave the following percentage composition by mass:H=9.09;O=36.36;C=54.55.Mol mass of compound is 88.Find it's molecular formula
Answer» Answer: The molecular formula will be $C_4H_8O_2$ Explanation: If percentage are given then we are taking TOTAL mass is 100 grams. So, the mass of each element is equal to the percentage given. Mass of C= 54.55 g Mass of H = 9.09 g Mass of O = 36.36 g STEP 1 : convert given masses into moles. Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{54.55g}{12g/mole}=4.5moles$ Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{9.09g}{1g/mole}=9.09moles$ Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{36.36g}{16g/mole}=<klux>2</klux>.3moles$ Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated. For C = $\frac{4.5}{2.3}=2$ For H = $\frac{9.09}{2.3}=4$ For O = $\frac{2.3}{2.3}=1$ The ratio of C : H: O= 2: 4:1 Hence the empirical formula is $C_2H_4O$ The empirical weight of $C_2H_4O$ = 2(12)+4(1)+1(16)= 44g. The molecular weight = 88 g/mole Now we have to calculate the molecular formula. $n=\frac{\text{Molecular weight of metal}}{\text{Equivalent of metal}}=\frac{88}{44}=2$ $2\times C_2H_4O=C_4H_8O_2$ The molecular formula will be=

a compound on analysis gave the following percentage composition by mass:H=9.09;O=36.36;C=54.55.Mol mass of compound is 88.Find it's molecular formula

Answer»

Answer: The molecular formula will be $C_4H_8O_2$

Explanation:

If percentage are given then we are taking TOTAL mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 54.55 g

Mass of H = 9.09 g

Mass of O = 36.36 g

STEP 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{54.55g}{12g/mole}=4.5moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{9.09g}{1g/mole}=9.09moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{36.36g}{16g/mole}=<klux>2</klux>.3moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{4.5}{2.3}=2$

For H = $\frac{9.09}{2.3}=4$

For O = $\frac{2.3}{2.3}=1$

The ratio of C : H: O= 2: 4:1

Hence the empirical formula is $C_2H_4O$

The empirical weight of $C_2H_4O$ = 2(12)+4(1)+1(16)= 44g.

The molecular weight = 88 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight of metal}}{\text{Equivalent of metal}}=\frac{88}{44}=2$

$2\times C_2H_4O=C_4H_8O_2$ The molecular formula will be=