5.34g of salt of formula M2SO4 (M is a metal)were dissolved in water. The sulfate ion was precipitated by adding excess barium chloride solution when 4.66g of barium sulfate (BaSO4) were obtained.

a)How many moles of sulfate ion were precipitated as barium sulfate?

b)How many moles of M2SO4 were in the solution?

c)What is the formula mass of M2SO4?

d)What is the RAM of M? Identify M.

a. mols = grams/formula mass BaSO4 = about 4.66/233 = estimated 0.02

b. mols M2SO4 = mols BaSO4 since
M2SO4 ==> BaSO4

c. mols = grams/formula mass so
formula mass = grams/mols = about 5.34/0.02 = about 267.

d. If M2SO4 = 267 then
2M must be 267-formula mass SO4 = about 267-96 = about 171 so M = about 171/2 = about 85.5 Look that up on the periodic table.

a) Well, let me do some calculations while I juggle some numbers. The molar mass of barium sulfate (BaSO4) is 233.37 g/mol. So, to find the number of moles, we divide the mass of barium sulfate (4.66g) by the molar mass:

moles of BaSO4 = 4.66g / 233.37 g/mol = 0.02 mol

Now, keep in mind that barium sulfate contains one sulfate ion (SO4) per formula unit. So, the number of moles of sulfate ions precipitated is the same as the number of moles of barium sulfate:

moles of sulfate ions = 0.02 mol

b) To determine the moles of M2SO4, we need to compare the moles of sulfate ions to the ratio of sulfate ions to M2SO4. From the formula, we know that one formula unit of M2SO4 contains one sulfate ion:

moles of M2SO4 = 0.02 mol

c) The formula mass of M2SO4 can be calculated by adding up the masses of each element in the formula. The sulfate ion (SO4) has a molar mass of 96.06 g/mol. And since there are two M's in M2SO4, we need to multiply the molar mass of M2SO4 by 2:

formula mass of M2SO4 = 2 * (molar mass of M) + molar mass of SO4

d) Unfortunately, I don't have enough information to determine the exact molar mass or identity of M. Without more details, I'll just have to take a wild guess and say that M is the element of surprise! So, let's go with that. M is the element of surprise, and its exact molar mass remains a mystery.

a) To find the moles of sulfate ion precipitated as barium sulfate, we need to use the molar mass of barium sulfate (BaSO4).

The molar mass of BaSO4 is the sum of the atomic masses:
1 atom of Ba (barium) = 137.33 g/mol
1 atom of S (sulfur) = 32.06 g/mol
4 atoms of O (oxygen) = 4 * 16.00 g/mol = 64.00 g/mol

Molar mass of BaSO4 = 137.33 + 32.06 + 64.00 = 233.39 g/mol

Now we can calculate the moles of barium sulfate precipitated:
moles of BaSO4 = mass of BaSO4 / molar mass of BaSO4
moles of BaSO4 = 4.66 g / 233.39 g/mol = 0.019988 mol (rounded to 4 decimal places)

Since the molar ratio between sulfate ion (SO4^2-) and barium sulfate (BaSO4) is 1:1, the number of moles of sulfate ion precipitated is also 0.019988 mol.

b) The number of moles of M2SO4 can be determined using the molar mass of M2SO4.

To find the molar mass of M2SO4, we need to know the formula mass of M2SO4:

c) The formula mass of M2SO4 is the sum of the atomic masses of M, S, and 4 O atoms.

Let's assume the formula mass of M2SO4 is X g/mol.

The molar mass of M2SO4 is then 2 times the atomic mass of M plus the atomic mass of S plus 4 times the atomic mass of O:

X = 2 * atomic mass of M + atomic mass of S + 4 * atomic mass of O

Unfortunately, without knowing the atomic mass of M or the identity of M, we cannot determine the formula mass of M2SO4 or answer part c).

d) Without knowing the RAM (Relative Atomic Mass) of M or its identity, we cannot determine the individual atomic mass or the identity of M. Therefore, we cannot answer part d) either.

a) To find the number of moles of sulfate ion precipitated as barium sulfate, you need to convert the mass of barium sulfate obtained (4.66g) to moles. The molar mass of BaSO4 can be found by summing the atomic masses of all its constituent elements:

Molar mass of BaSO4 = (atomic mass of Ba) + (atomic mass of S) + 4 * (atomic mass of O)

Let's calculate the molar mass of BaSO4:

Molar mass of Ba = 137.33 g/mol (from the periodic table)
Molar mass of S = 32.06 g/mol (from the periodic table)
Molar mass of O = 16.00 g/mol (from the periodic table)

Molar mass of BaSO4 = 137.33 + 32.06 + 4 * 16.00 = 233.39 g/mol

Now, we can calculate the number of moles of BaSO4:

Moles of BaSO4 = Mass of BaSO4 / Molar mass of BaSO4

Moles of BaSO4 = 4.66g / 233.39 g/mol = 0.020 moles

Since BaSO4 has a 1:1 stoichiometric ratio with sulfate ions (SO4^2-), the number of moles of sulfate ion precipitated is also 0.020 moles.

b) Since the formula of the salt is M2SO4, it indicates that there are 2 moles of M for every mole of sulfate ion (SO4^2-) in the compound. Therefore, the number of moles of M2SO4 in the solution is twice the number of moles of sulfate ion precipitated.

Moles of M2SO4 = 2 * moles of sulfate ion

Moles of M2SO4 = 2 * 0.020 moles = 0.040 moles

c) The formula mass of M2SO4 can be calculated by summing the molar masses of all its constituent elements according to the formula:

Formula mass of M2SO4 = (2 * atomic mass of M) + (atomic mass of S) + 4 * (atomic mass of O)

Unfortunately, without knowing the identity of the metal (M), we cannot directly calculate the formula mass of M2SO4.

d) To determine the relative atomic mass (RAM) of M, we would need additional information about the compound or the metal itself. Without this information, we cannot identify the metal (M) or calculate its relative atomic mass.