A compound is made up of two elementsA and B,has A=70%,B=30%.The relative number of moles of Aand B in the compound are 1.25 and 1.88 respectivily.If the molecular mass of the compound is 160g.Find the molecular formula of the compound?
To find the molecular formula of the compound, we need to determine the empirical formula first.
1. Convert the given percentages to masses:
Mass of A = 70% x 160g = 112g
Mass of B = 30% x 160g = 48g
2. Calculate the moles of each element:
Moles of A = Mass of A / Atomic Mass of A
= 112g / Atomic Mass of A
Moles of B = Mass of B / Atomic Mass of B
= 48g / Atomic Mass of B
3. Calculate the ratio of moles of A to B:
Ratio (A to B) = Moles of A / Moles of B
= (112g / Atomic Mass of A) / (48g / Atomic Mass of B)
4. Given that the relative number of moles of A and B in the compound are 1.25 and 1.88 respectively, set up the equation:
1.25 = (112g / Atomic Mass of A) / (48g / Atomic Mass of B)
1.88 = (112g / Atomic Mass of A) / (48g / Atomic Mass of B)
5. Simplify the equation by removing the common factors:
1.25 = (112 / Atomic Mass of A) / (48 / Atomic Mass of B)
1.88 = (112 / Atomic Mass of A) / (48 / Atomic Mass of B)
6. Cross-multiply and solve for Atomic Mass of A:
1.25 x (48 / Atomic Mass of B) = 112 / Atomic Mass of A
1.25 x 48 = (112 / Atomic Mass of A) x Atomic Mass of B
Replace (112 / Atomic Mass of A) with (1.88 x (48 / Atomic Mass of B)):
1.25 x 48 = (1.88 x (48 / Atomic Mass of B)) x Atomic Mass of B
Simplify:
60 = (1.88 x 48) / (Atomic Mass of A)
(Atomic Mass of A) = (1.88 x 48) / 60
7. Substitute the value of Atomic Mass of A back into one of the equations given in step 4 to solve for Atomic Mass of B.
8. Once you have the atomic masses of both elements, find their respective atomic masses using the periodic table.
9. Divide the atomic masses by the smallest atomic mass value obtained to obtain the simplest ratio of atoms.
10. Round off the ratio to the nearest whole number to get the empirical formula.
Now, with the empirical formula, you can find the molecular formula by finding the multiple by which the empirical formula must be multiplied to give the molecular mass (as given in the problem, 160g).
To find the molecular formula of the compound, we need to determine the actual number of moles of each element in the compound.
Let's assume that the compound contains 1 mole of both elements A and B. In that case, the relative number of moles of A and B would be equal to 1.
However, in this case, the relative number of moles of A is given as 1.25, and the relative number of moles of B is given as 1.88. This means that the actual number of moles of A and B in the compound is higher than 1.
To calculate the actual number of moles of A and B, we can use the following formulas:
moles of A = (relative number of moles of A) x (moles of B in 1 mole of compound)
moles of B = (relative number of moles of B) x (moles of A in 1 mole of compound)
First, let's find the moles of B in 1 mole of compound. Since the compound is made up of 30% B, we can calculate it as follows:
moles of B in 1 mole of compound = 0.30
Now, substituting the values into the formulas, we have:
moles of A = 1.25 x (0.30) = 0.375
moles of B = 1.88 x (1) = 1.88
Next, we need to find the molar masses of elements A and B. Since the molecular mass of the compound is given as 160 g, we can use the percentages of A and B to calculate their molar masses.
molar mass of A = (A percentage / 100) x (molecular mass of compound) = (0.70) x (160 g) = 112 g
molar mass of B = (B percentage / 100) x (molecular mass of compound) = (0.30) x (160 g) = 48 g
Now, we can calculate the actual masses of A and B in the compound:
mass of A = (moles of A) x (molar mass of A) = 0.375 x 112 g = 42 g
mass of B = (moles of B) x (molar mass of B) = 1.88 x 48 g = 90.24 g
Finally, we can calculate the ratio of the masses of A and B:
mass ratio of A to B = (mass of A) / (mass of B) = 42 g / 90.24 g
Simplifying this ratio, we find that A is to B in a approximately 0.466:1 ratio.
Since the molecular formula of the compound is determined by the ratio of the number of atoms of each element, we can conclude that the molecular formula of the compound is approximately A₂B.