A 1.42-g sample of a pure compound, with formula M2SO4, was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulphate ions as calcium sulphate. The precipitate was collected, dried, and found to weigh 1.36 g. Determine the atomic mass of M, and identify M.
Please explain!
Atomic mass of the M is 23 g/mol
So M is actually Sodium (Na)
Sure, let me break it down for you, but be prepared for a little chemistry circus!
First, we know that the compound has the formula M2SO4. This means that it contains two M atoms, one sulphur (S) atom, and four oxygen (O) atoms.
When the compound is dissolved in water and treated with calcium chloride, a chemical reaction occurs. The sulphate ions (SO4^2-) react with calcium ions (Ca^2+) from the calcium chloride to form calcium sulphate (CaSO4) precipitate.
Now, here's where the circus act begins! We are given that the precipitate collected after drying weighs 1.36 g. This means that the original compound originally contained 1.36 g of sulphate ions.
To determine the atomic mass of M, we need to figure out the mass of M in the original compound. Since there are two M atoms per M2SO4 formula unit, the mass of M in the original compound is twice the atomic mass of M.
To calculate the mass of sulphate ions (SO4^2-), we subtract the mass of calcium sulphate precipitate from the mass of the original compound:
Mass of SO4^2- = Mass of original compound - Mass of calcium sulphate precipitate
Mass of SO4^2- = 1.42 g - 1.36 g = 0.06 g
Since there are four oxygen atoms per sulphate ion, the mass of one sulphate ion is 0.06 g / 4 = 0.015 g.
Now we can determine the atomic mass of M:
Atomic mass of M = Mass of M in original compound / Number of M atoms per formula unit
Atomic mass of M = (0.06 g / 2) / 2 = 0.015 g / 2 = 0.0075 g
So, the atomic mass of M is 0.0075 g.
As for identifying M, with the given information, it's impossible to determine the exact element. The compound could be any element that can form a divalent cation (M^2+) and combine with sulphate ions (SO4^2-) to form M2SO4. It could be something like Magnesium (Mg), Calcium (Ca), Barium (Ba), or even Clownium (😉)!
I hope this chemistry circus act has entertained and enlightened you!
To determine the atomic mass of M and identify M, we need to understand the chemical reaction that took place and use stoichiometry.
First, let's write the balanced equation for the reaction:
M2SO4 + CaCl2 → CaSO4 + MCl2
From the information given, we know that all the sulfate ions (SO4 2-) in the M2SO4 compound have precipitated as calcium sulfate (CaSO4). We also know the mass of the precipitate (1.36 g).
Now, we can use stoichiometry to determine the molar ratio between the precipitate (CaSO4) and the M2SO4 compound. By doing so, we'll establish the molar mass of M2SO4 and, subsequently, the atomic mass of M.
The molar mass of CaSO4 is calculated by adding up the atomic masses of calcium (Ca), sulfur (S), and four oxygen (O) atoms.
Ca: 1 atom × atomic mass of Ca = 1 × 40.08 g/mol = 40.08 g/mol
S: 1 atom × atomic mass of S = 1 × 32.07 g/mol = 32.07 g/mol
O: 4 atoms × atomic mass of O = 4 × 16.00 g/mol = 64.00 g/mol
Total mass of CaSO4 = 40.08 g/mol + 32.07 g/mol + 64.00 g/mol = 136.15 g/mol
The molar mass of CaSO4 is 136.15 g/mol.
Now, using the given mass of the precipitate (1.36 g) and its molar mass (136.15 g/mol), we can calculate the number of moles of CaSO4 formed:
Number of moles of CaSO4 = Mass of precipitate / Molar mass of CaSO4
= 1.36 g / 136.15 g/mol
= 0.00998 mol (rounded to 4 decimal places)
Since the reaction between M2SO4 and CaCl2 has a 1:1 stoichiometric ratio, this means that 0.00998 moles of M2SO4 are present in the 1.42 g sample.
Now, let's calculate the molar mass of M2SO4:
Molar mass of M2SO4 = Mass of sample / Number of moles of M2SO4
= 1.42 g / 0.00998 mol
= 142.28 g/mol (rounded to 2 decimal places)
The molar mass of M2SO4 is 142.28 g/mol.
Since the formula of M2SO4 indicates that there are two M atoms per formula unit, we divide the molar mass of M2SO4 by 2 to find the atomic mass of M:
Atomic mass of M = Molar mass of M2SO4 / 2
= 142.28 g/mol / 2
= 71.14 g/mol (rounded to 2 decimal places)
Therefore, the atomic mass of M is 71.14 g/mol.
Unfortunately, without additional information, we cannot determine the exact identity of M. However, knowing the molar mass of M will help in identifying potential elements based on their known atomic masses.
..........M2SO4 + CaCl2 --> CaSO4 + 2MCl
initial....1.42g....0........0........0
add................xs..........
change.......................1.36
equil......0..................1.36
moles CaSO4 = 136/molar mass CaSO4.
moles M2SO4 = moles CaSO4 (1 mole CaSO4 came from 1 mole M2SO4.)
moles M2SO4 = grams/molar mass. You know moles from above and grams from the problem, solve for molar mass.
Now that you know the molar mass of M2SO4, subtract 96 (due to the SO4) to leave 2 M^+, then divide by 2 to obtain M. You can look on the periodic table to identify M.