A sample of silver chromate (Ag2CrO4) has a mass of 29.0 g. How many CrO42- ions are present?
Same method as above. moles = gram/molar mass. There is 1 CrO4^-2 per mole of Ag2CrO4.
To determine the number of CrO42- ions present in the sample of silver chromate (Ag2CrO4), you need to calculate the number of moles of the compound and then use the mole ratio to find the number of CrO42- ions.
Here's how to do it step by step:
1. Find the molar mass of Ag2CrO4:
- Molar mass of Ag = 107.87 g/mol (atomic mass of silver)
- Molar mass of Cr = 52.00 g/mol (atomic mass of chromium)
- Molar mass of O = 16.00 g/mol (atomic mass of oxygen)
- Molar mass of Ag2CrO4 = (2 * 107.87) + 52.00 + (4 * 16.00) = 331.87 g/mol
2. Calculate the number of moles of Ag2CrO4:
- Moles = Mass / Molar mass
- Moles = 29.0 g / 331.87 g/mol ≈ 0.0874 mol
3. Use the mole ratio to find the number of CrO42- ions:
- From the chemical formula Ag2CrO4, we can see that there is one CrO42- ion for every Ag2CrO4 ion.
- Therefore, the number of CrO42- ions is equal to the number of Ag2CrO4 ions in the sample, which is equal to the number of moles of Ag2CrO4.
So, in this case, there are approximately 0.0874 mol of CrO42- ions present in the sample of silver chromate.