on complete combustion of one mole of a certain alkane , two moles of C02 and three moles of water was produced. determine the relative molecular mass of the alkane.
(C= 12, H=1)
To determine the relative molecular mass of the alkane, we need to gather some information about the products formed during its combustion.
First, we know that two moles of carbon dioxide (CO2) and three moles of water (H2O) are produced from the combustion of one mole of the alkane. Second, we are given the atomic masses of carbon (C = 12) and hydrogen (H = 1).
Since each molecule of carbon dioxide (CO2) contains one carbon atom, and each molecule of water (H2O) contains two hydrogen atoms, we can calculate the total number of carbon and hydrogen atoms in the products.
Number of carbon atoms:
2 moles of CO2 x 1 mole of C/1 mole of CO2 = 2 moles of C
Number of hydrogen atoms:
3 moles of H2O x 2 moles of H/1 mole of H2O = 6 moles of H
Now, we can find the molecular formula of the alkane using the number of carbon and hydrogen atoms present in the products.
For every two moles of carbon in the products, we need to have two moles of carbon in the alkane.
For every six moles of hydrogen in the products, we need to have twelve moles of hydrogen in the alkane.
Therefore, the molecular formula of the alkane is C2H12.
To calculate the relative molecular mass of the alkane, we sum up the atomic masses of all the atoms in the molecular formula.
Relative molecular mass of the alkane:
(2 x atomic mass of carbon) + (12 x atomic mass of hydrogen)
= (2 x 12) + (12 x 1)
= 24 + 12
= 36
Therefore, the relative molecular mass of the alkane is 36.
To determine the relative molecular mass of the alkane, we need to consider the balanced equation for the combustion reaction.
The balanced equation for the complete combustion of the alkane can be written as follows:
CₙH₂ₙ₊₂ + (3n + 1/2) O₂ → n CO₂ + (n + 1) H₂O
From the given information, we know that two moles of CO₂ and three moles of H₂O are produced. So, we can set up the following ratios:
2 moles CO₂ / n moles alkane = moles CO₂ per mole of alkane
3 moles H₂O / n moles alkane = moles H₂O per mole of alkane
Now, let's substitute the values into the equation:
2 / n = n CO₂ per mole of alkane
3 / n = n + 1 H₂O per mole of alkane
Since we know that two moles of CO₂ and three moles of H₂O are produced, we can equate those ratios:
2 / n = n + 1 / 3
To solve for n, we can cross-multiply:
2 * 3 = (n + 1) * n
6 = n² + n
n² + n - 6 = 0
Factoring or using the quadratic formula, we find that n = 2 or n = -3. However, since n represents the number of carbon atoms in the alkane, it cannot be negative. Therefore, n = 2.
Now, we can calculate the relative molecular mass (Mr) of the alkane:
Mr = (n * atomic mass of C) + ((2n + 2) * atomic mass of H)
Substituting the values:
Mr = (2 * 12) + ((2 * 2 + 2) * 1)
Mr = 24 + 6
Mr = 30
Therefore, the relative molecular mass of the alkane is 30.
30g
CxHy + O2 --> 2CO2 + 3H2O
So x must be 2 and y must be 6; i.e., C2H6
Molar mass C2H6 is ............