A sample of silver chromate (Ag2CrO4) has a mass of 29.0 g.
(a) How many Ag+ ions are present?
(b)What is the mass in grams of one formula unit of silver chromate?
(c) How many CrO42- ions are present?
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To answer these questions, we first need to know the molar mass of silver chromate (Ag2CrO4).
(a) To determine the number of Ag+ ions present, we need to consider that there are two Ag+ ions per one Ag2CrO4 formula unit.
To find the number of moles of Ag2CrO4, we need to divide the given mass (29.0 g) by its molar mass. The molar mass of Ag2CrO4 is calculated as follows:
Ag2CrO4:
- Two silver atoms (Ag) with an atomic mass of 107.87 g/mol each
- One chromium atom (Cr) with an atomic mass of 52.00 g/mol
- Four oxygen atoms (O) with an atomic mass of 16.00 g/mol each
Calculating the molar mass of Ag2CrO4:
(2 * 107.87 g/mol) + 52.00 g/mol + (4 * 16.00 g/mol) = 331.74 g/mol
Now we can calculate the number of moles of Ag2CrO4:
29.0 g / 331.74 g/mol = 0.0874 mol
Since there are two Ag+ ions per formula unit, we can multiply the number of moles of Ag2CrO4 by 2:
0.0874 mol * 2 = 0.1748 mol
Therefore, there are 0.1748 moles of Ag+ ions present.
To convert this to the number of Ag+ ions, we use Avogadro's number (6.022 x 10^23):
0.1748 mol * (6.022 x 10^23 ions/mol) = 1.05 x 10^23 Ag+ ions
So, there are approximately 1.05 x 10^23 Ag+ ions present.
(b) The mass of one formula unit of silver chromate can be calculated by simply dividing the molar mass of Ag2CrO4 by Avogadro's number:
331.74 g/mol / (6.022 x 10^23/mol) ≈ 5.50 x 10^-22 g
Thus, one formula unit of silver chromate has a mass of approximately 5.50 x 10^-22 grams.
(c) Similarly, to determine the number of CrO4^2- ions present, we need to note that there is one CrO4^2- ion per one Ag2CrO4 formula unit.
Using the same number of moles as calculated in part (a) (0.0874 mol), we can say that there are 0.0874 moles of CrO4^2- ions.
To convert this to the number of CrO4^2- ions, again using Avogadro's number:
0.0874 mol * (6.022 x 10^23 ions/mol) = 5.27 x 10^22 CrO4^2- ions
Hence, there are approximately 5.27 x 10^22 CrO4^2- ions present.