I completed this question like you told me to. Are the answers correct? Is my method for finding the molecular formula correct?
2.4g of a compound of carbon, hydrogen and oxygen gave on combustion, 3.52g of CO2 and 1.44g of H2O. The relative molecular mass of the compound was found to be 60.
a)What are the masses of carbon, hydrogen and oxygen in 2.4g of the compound?
b)What are the emperical and molecular formulae of the compound?
Ans:
a) 3.52g CO2 x (atomic mass C/molar mass CO2) = 0.96g C.
1.44g H2O x (2*atomic mass H/molar mass H2O)= 0.16g H.
g of O = 2.4 - gC - g H
2.4 - 0.96- 0.16= 1.28g O
b) e.f
C H O
0.96/12 0.16/1 1.28/16
mol=0.08 0.16 0.08
e.f= 1 2 1
emperical formula= CH2O
m.f
(CH2O)n= 60 **60/30=2**
molecular formula= C2H4O2
All of your work looks good to me.
A 0.039 M solution of a weak acid (HA) has a pH of 4.28. What is the Ka of the acid?
To verify if the answers are correct, let's go through the method of finding the molecular formula again.
a) To calculate the mass of each element in 2.4g of the compound, we need to determine the mass of carbon, hydrogen, and oxygen separately.
First, let's find the mass of carbon (C):
The equation for the combustion of carbon to form carbon dioxide (CO2) is:
C + O2 -> CO2
From the given data, we know that 3.52g of CO2 is produced. Using the molar mass of CO2, which is 44 g/mol, we can calculate the mass of carbon:
3.52g CO2 x (atomic mass C/molar mass CO2) = 3.52g CO2 x (12g C/44g CO2) = 0.96g C
So, the mass of carbon in 2.4g of the compound is 0.96g.
Next, let's find the mass of hydrogen (H):
The equation for the combustion of hydrogen to form water (H2O) is:
2H2 + O2 -> 2H2O
From the given data, we know that 1.44g of H2O is produced. Using the molar mass of H2O, which is 18 g/mol, we can calculate the mass of hydrogen:
1.44g H2O x (2*atomic mass H/molar mass H2O) = 1.44g H2O x (2*1g H/18g H2O) = 0.16g H
So, the mass of hydrogen in 2.4g of the compound is 0.16g.
Now, let's find the mass of oxygen (O):
To calculate the mass of oxygen, subtract the masses of carbon and hydrogen from the total mass of the compound (2.4g):
g of O = 2.4g - gC - gH = 2.4g - 0.96g - 0.16g = 1.28g O
So, the mass of oxygen in 2.4g of the compound is 1.28g.
b) Now, let's determine the empirical and molecular formulas of the compound.
For the empirical formula, divide the masses of each element by their respective atomic masses and then divide those values by the smallest value obtained:
Empirical formula:
C: 0.96/12 = 0.08
H: 0.16/1 = 0.16
O: 1.28/16 = 0.08
Dividing each value by 0.08, we get:
C: 0.08/0.08 = 1
H: 0.16/0.08 = 2
O: 0.08/0.08 = 1
Therefore, the empirical formula is CH2O.
To determine the molecular formula, you need the relative molecular mass of the compound, which is given as 60. Divide the molar mass of the empirical formula (CH2O) by the empirical formula mass (30) to find the value of "n":
60/30 = 2
This means that the molecular formula is (CH2O)n, where "n" is 2. Therefore, the molecular formula is C2H4O2.
Based on these calculations, the answers provided are correct.