A sample of a substance containing only C and H burns in excess O₂ to yield 4.4g of CO₂
and 2.7 g of H₂O. The empirical formula of a substance is?

(C = 12, O = 16, H = 1)

VPETERS2002

6 years ago

For more explanation:

From the question above, Carbon (C) burns in excess oxygen (2O2) to yield 4.4g of CO2
Chemical reaction: C+O2----CO2
since Oxygen is in excess amount (2O2)
Balane Chemical reaction: 2C+2O2----2CO2
where;
Molar mass of 2C= 2*12=24g
Molar mass of 2CO2=(2*12)+(2*16*2)=88g
So therefore,
88g of CO2 was produced by 24g of 2C
4.4g of CO2 will be produced by=(4.4*24)/88=1.2g of 2C

Also from the question above, Hydrogen (H) burns in excess oxygen (2O2) to yield 2.7g of H2O
Chemical reaction: H2+O----H2O
since Oxygen is in excess amount (2O2)
Balane Chemical reaction: 8H+2O2----4H2O
where;
Molar mass of 8H=8*1=8g
Molar mass of 4H2O=(4*1*2)+(4*16)=72g
So therefore,
72g of 4H2O was produced by 8g of 8H
2.7g of 4H2O will be produced by=(2.7*8)/72=0.3g of 8H

Hence, mass of carbon=1.2g and mass of hydrogen=0.3g

olyezema

4 years ago

x=Ar×m/M
X=mass composition of the element
Ar=R.M.M of the element
m= mass if the product
M=R.M.M of the product
C=12/44 ×4.4=1.2g
H=2/18×2.7=0.3g
C
1.2/12
0.1/0.1=1
H
0.3/1
0.3/0.1=3
CH3

Ogbonna_Chisom

3 years ago

44g of CO2​→1mol of carbon

∴4.4gofCO2​→4.4/44=0.1mol of C

18g of H2​O→2mol of Hydrogen

∴2.7g of H2​O→2×2.7/18=0.3mol of H

molar ratio of C:H is 0.1:0.3

Divi​ding by the smallest which is 0.1, we get CH3

Som15432

1 year ago

Here's how to figure out the empirical formula of the substance:
1. Find the moles of carbon (C) and hydrogen (H):
* Moles of CO2:
* The molar mass of CO2 is about 44 g/mol.
* Moles of CO2 = 4.4 g / 44 g/mol = 0.1 mol
* Since each CO2 molecule has 1 carbon atom, there are 0.1 moles of carbon.
* Moles of H2O:
* The molar mass of H2O is about 18 g/mol.
* Moles of H2O = 2.7 g / 18 g/mol = 0.15 mol
* Since each H2O molecule has 2 hydrogen atoms, there are 0.15 mol * 2 = 0.3 moles of hydrogen.
2. Find the mole ratio of C to H:
* Divide both mole values by the smallest one (0.1 mol):
* C: 0.1 mol / 0.1 mol = 1
* H: 0.3 mol / 0.1 mol = 3
3. Write the empirical formula:
* The ratio of C to H is 1:3, so the empirical formula is CH3.
Therefore, the empirical formula of the substance is CH3.

solgade

1 year ago

For C:

mass of C in 4.4g of CO2 is

C = [12/44] × 4.4
C = 1.2

For H:

Mass of H in H20 is

H =[2/18] x 2.7
H = 0.3

C || H
1.2 || 0.3
÷ 12 || ÷ 1

0.1 || 0.3
÷ 0.1 || ÷ 0.1

1 || 3

CH3