Answer:
The molecular formula for the compound is Ag₂C₂O₄, silver oxalate.
Explanation:
This is an easy excersise, we relate the moles of the elements in the compound with the molar mass of the total compound.
The sample weighs 1.372 g, with 0.288 g of O, 0.974 g of Ag and the rest, C
Therefore, 1.372g - 0.288 g - 0.974g = 0.112 g are C
We convert the mass to moles
0.288 g / 16g/mol = 0.018 moles of O
0.974 g / 107.87 g/mol = 0.00903 moles of Ag
0.112 g / 12 g/mol = 0.00933 moles of C
Now, the use of the molar mass of the compound. Let's prepare some rules of three:
1.372 g of compound contain 0.018 moles of O, 0.00903 moles of Ag and 0.00933 moles of C
Then, 308.8 g of compound (a mol) must contain:
(308.8 . 0.018) / 1.372 = 4 moles of O
(308.8 . 0.00903) / 1.372 = 2 moles of Ag
(308.8 . 0.00933) / 1.372 = 2 moles of C
The molecular formula for the compound is Ag₂C₂O₄, silver oxalate.
Answer:
The molecular formula is Ag2C2O4
Explanation:
Step 1: Data given
A compound contains C, O and Ag
The mass of the compound is 1.372 grams
Mass of O = 0.288 grams
Mass of Ag = 0.974 grams
Molar mass C = 12.01 g/mol
Molar mass O = 16.0 g/mol
Molar mass Ag =107.87 g/mol
The molar mass of the compound = 308.8 g/mol
Step 2: Calculate moles
Moles = mass / molar mass
Moles O = 0.288 grams / 16.0 grams
Moles O = 0.018 moles
Moles Ag = 0.974 grams / 107.87 g/mol
Moles Ag = 0.00903 moles
Moles C = 0.11 grams / 12.01 g/mol
Moles C = 0.00916 moles
Step 3: Calculate the mol ratio
We divide by the smallest amount of moles
O: 0.018/ 0.00903 = 2
C: 0.0.00916 / 0.00903 = 1
Ag: 0.00903/0.00903 =
The empirical formula is AgCO2
The molar mass of this empirical formula is 151.88
We have to multiply the empirical formula by n
n = 308.8/151.88 = 2
2* (AgCO2) = Ag2C2O4
The molecular formula is Ag2C2O4
