0.3975 g of an oxide of divalent metal on reduction with hydrogen gives 0.090 g of water. The atomic weight of metal is:
a) 63.5
b) 24
c) 55.5
d) 40
MO + H2 ==>M + H2O
0.090 x (molar mass MO/molar mass H2O) = 0.3975
Solve for molar mass MO, subtract 18 to arrive at atomic mass M.
To find the atomic weight of the metal, we need to use the concept of stoichiometry and conversion factors.
Step 1: Find the moles of water produced.
We know that the mass of water produced is 0.090 g. To convert this mass to moles, we need to divide it by the molar mass of water.
The molar mass of water (H2O) is: 2(1.008 g/mol) + 16.00 g/mol = 18.016 g/mol.
Moles of water = Mass of water / Molar mass of water
= 0.090 g / 18.016 g/mol
≈ 0.00499 mol
Step 2: Use the balanced chemical equation to establish the mole ratio between water and the oxide.
The balanced chemical equation is:
Metal Oxide + Hydrogen (H2) → Metal + Water (H2O)
From the balanced equation, we can see that the mole ratio between the oxide and water is 1:1. This means that for every mole of oxide, we get one mole of water.
Step 3: Calculate the moles of the oxide of the divalent metal.
Since the mole ratio is 1:1, the moles of the oxide of the divalent metal are also approximately 0.00499 mol.
Step 4: Use the concept of atomic weight.
Atomic weight (also called atomic mass) is the mass of an individual atom of an element compared to 1/12th the mass of an atom of carbon-12.
To find the atomic weight of the metal, we need to divide the mass of the oxide of the divalent metal by the number of moles of the oxide of the divalent metal.
Atomic weight of metal = Mass of oxide of divalent metal / Moles of oxide of divalent metal
= 0.3975 g / 0.00499 mol
≈ 79.558 g/mol
Now, we need to find the element with an atomic weight of approximately 79.558 g/mol from the answer choices provided:
a) 63.5 g/mol
b) 24 g/mol
c) 55.5 g/mol
d) 40 g/mol
The only option close to 79.558 g/mol is option a) 63.5 g/mol.
Therefore, the atomic weight of the metal is approximately 63.5.
To determine the atomic weight of the metal, we can use the concept of stoichiometry and the balanced chemical equation for the reduction of the oxide using hydrogen.
Let's assume the oxide of the divalent metal is MxOy. The reaction can be represented as:
MxOy + y/2 H2 → xM + y/2 H2O
From the given information, we know that 0.3975 g of the oxide gives 0.090 g of water.
Thus, we need to find the molar mass of the metal, which is equal to x times the atomic weight of the metal (M).
To find the molar mass of the water formed, we use the formula:
molar mass of water (H2O) = 2(H) + 16(O) = 18 g/mol
Now, we can set up a proportion to relate the mass of the oxide and the mass of the water formed:
(0.3975 g of MxOy) / (molar mass of MxOy) = (0.090 g of H2O) / (molar mass of H2O)
Since the ratio of Molar mass of MxOy to the molar mass of H2O is x to y/2, we can rewrite the proportion as:
(0.3975 g of MxOy) / (xM) = (0.090 g of H2O) / (18 g/mol)
Cross-multiplying the equation, we get:
(0.3975 g of MxOy) * (18 g/mol) = (0.090 g of H2O) * (xM)
0.3975 * 18 = 0.090x
Simplifying this equation, we find:
7.155 = 0.090x
Finally, solving for x:
x = 7.155 / 0.090
x ≈ 79.5
Therefore, the approximate atomic weight of the metal (M) is 79.5 g/mol.
Looking at the answer choices provided, the closest approximation to 79.5 is 63.5, so the correct answer is (a) 63.5.