Use the Ar values below to calculate: (Ar values: Fe = 55.80, B = 10.80, O = 16.00) 1-the relative molecular mass of iron(III) borate, Fe(BO2 )3 . 2- the number of moles in 92.1g of iron(III) borate
The Correct Answer and Explanation is:
Let’s solve each part step by step using the given Ar values:
- Fe = 55.80
- B = 10.80
- O = 16.00
1. Relative Molecular Mass of Iron(III) Borate, Fe(BO₂)₃
Iron(III) borate has the chemical formula Fe(BO₂)₃.
We break this compound down:
- 1 Fe atom
- 3 BO₂ groups
- Each BO₂ group has 1 B and 2 O atoms
So, in total:
- 1 Fe
- 3 × 1 = 3 B
- 3 × 2 = 6 O
Now multiply the Ar (atomic relative mass) by the number of atoms:
- Fe: 1 × 55.80 = 55.80
- B: 3 × 10.80 = 32.40
- O: 6 × 16.00 = 96.00
Add them all together:
Relative molecular mass = 55.80 + 32.40 + 96.00 = 184.20 g/mol
2. Number of Moles in 92.1 g of Iron(III) Borate
Use the formula: Moles=MassRelative Molecular Mass\text{Moles} = \frac{\text{Mass}}{\text{Relative Molecular Mass}}
Given:
- Mass = 92.1 g
- Relative Molecular Mass = 184.20 g/mol
Moles=92.1184.20≈0.500 mol\text{Moles} = \frac{92.1}{184.20} ≈ 0.500 \text{ mol}
Explanation
To find the relative molecular mass of a compound, we sum up the relative atomic masses (Ar) of each element, considering how many atoms of each are present. For iron(III) borate, the chemical formula is Fe(BO₂)₃. This means the compound consists of one iron (Fe) atom and three borate units (BO₂). Each BO₂ group contains one boron (B) and two oxygen (O) atoms.
By multiplying the Ar values with the number of respective atoms—Fe (55.80), B (10.80), and O (16.00)—we find the total relative molecular mass is 184.20 g/mol.
In part two, we apply a basic concept from chemistry: the mole. A mole is a standard unit for amount of substance, and one mole of a compound has a mass equal to its molecular mass in grams. To calculate how many moles are in a given mass (here, 92.1 g), we divide the mass by the relative molecular mass. Moles=Given MassMolar Mass=92.1184.20≈0.500 mol\text{Moles} = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{92.1}{184.20} ≈ 0.500 \text{ mol}
This tells us that 92.1 g of Fe(BO₂)₃ contains 0.500 moles of the compound. These calculations are foundational in chemistry, especially when converting between mass and amount in reactions, determining reagent quantities, or calculating yield.
