Magnesium carbonate (MgCO3) is a hydrate. If you start with 15.67 g of the hydrate and after heating you get 7.58 g, what is the formula of the hydrate? I'm having a lot of trouble on this one.
how many grams of water evaporated?
15.67 - 7.58 = 8.09 gramsH2O
how many mols of water is that
mol of water = 2 + 16 = 18 grams
so this is just about 8.09 / 18 = 0.45 mol of water
NOW how many mols of MgCO3 did you have
7.58 grams of MgCO3
mol = 24.3 + 12 + 3*16 = 84.3 grams
so you have 7.58/84.3 = 0.09 mol of MgCO3
what is the ratio of mols H2O to mols MgCO3 ?
.45 / .09 = 5 on the nose !
MgCO3 * 5H2O
To find the formula of the hydrate, we need to determine the number of water molecules associated with each formula unit of magnesium carbonate.
1. Calculate the mass of magnesium carbonate after heating:
Mass of magnesium carbonate before heating = 15.67 g
Mass of magnesium carbonate after heating = 7.58 g
Mass of water lost = Mass before heating - Mass after heating
= 15.67 g - 7.58 g
= 8.09 g
2. Convert the mass of water lost to moles:
Molar mass of water (H2O) = 18.015 g/mol
Moles of water lost = Mass of water lost / Molar mass of water
= 8.09 g / 18.015 g/mol
≈ 0.449 mol
3. Calculate the molar mass of magnesium carbonate:
Molar mass of magnesium (Mg) = 24.305 g/mol
Molar mass of carbon (C) = 12.011 g/mol
Molar mass of oxygen (O) = 16.00 g/mol
Molar mass of magnesium carbonate (MgCO3) = (24.305 g/mol) + (12.011 g/mol) + (3 × 16.00 g/mol)
≈ 84.313 g/mol
4. Determine the molar ratio of water to magnesium carbonate:
Moles of water lost / Moles of magnesium carbonate = 0.449 mol / 1 mol
≈ 0.449
5. The formula of the hydrate can be expressed as:
MgCO3 • xH2O
Since the molar ratio of water to magnesium carbonate is approximately 0.449, the formula of the hydrate is:
MgCO3 • 0.449H2O
To determine the formula of the hydrate, we need to find the molar mass of the anhydrous compound (MgCO3) and the molar mass of the water (H2O) lost during heating.
1. Calculate the molar mass of MgCO3 (anhydrous):
- Magnesium (Mg) has a molar mass of 24.31 g/mol.
- Carbon (C) has a molar mass of 12.01 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- Add them up: 24.31 + 12.01 + (3 x 16.00) = 84.32 g/mol.
2. Calculate the molar mass of H2O:
- Hydrogen (H) has a molar mass of 1.01 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- Add them up: (2 x 1.01) + 16.00 = 18.02 g/mol.
3. Determine the mass of the water lost by subtracting the final mass (7.58 g) from the initial mass (15.67 g):
- Mass of water lost = 15.67 g - 7.58 g = 8.09 g.
4. Calculate the number of moles of water lost by dividing the mass of water lost by its molar mass:
- Moles of water lost = 8.09 g / 18.02 g/mol ≈ 0.45 mol.
5. Calculate the number of moles of MgCO3 present in the sample by dividing its mass by its molar mass:
- Moles of MgCO3 = 7.58 g / 84.32 g/mol ≈ 0.09 mol.
6. Divide the moles of water lost and the moles of MgCO3 by the smallest number of moles to find the mole ratio:
- Moles ratio of water:MgCO3 = 0.45 mol / 0.09 mol ≈ 5:1.
7. Based on the mole ratio from step 6, we can determine the formula of the hydrate. The formula shows that there are 5 water molecules associated with 1 molecule of MgCO3. Therefore, the formula of the hydrate is MgCO3·5H2O.
So, the formula of the hydrate is MgCO3·5H2O.