Question

A compound has the formula XMnO4. X accounts for roughly 25% of the mass of the...

  1. A compound has the formula XMnO4. X accounts for roughly 25% of the mass of the compound.  
    1. Which is the cation and which is the anion in this compound, X or MnO4?
    2. What is the name of the anion?
    3. What is the charge of the anion?
    4. What is the likely group of the cation?
    5. Ions have a molecular mass the same way that atoms have an atomic mass. This is found by adding together the atomic masses of the constituents. For example, water, H2O has a formula mass of 18.(1H + 1H + 1O) or roughly (1+1+16). This is the mass of 1 mole of the compound. Given that, what is the mass of MnO4, in 3 significant figures?
    6. What is the mass of X in 3 significant figures?
    7. What is the symbol of X
    8. What is the name of this compound?
    9. What is the formula mass of this compound, in the appropriate number of significant figures
    10. If you have 50.5g of this compound, how many moles do you have?
    11. If dissolve 50.5 g of this compound into 500.0 mL of water, what is the concentration in moles/L
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Answer #1

a.

The compound is written as XMn04 . Since cations are always written before anions in ionic compound, X^+ is the cation and Mno is the anion.

b.

The name of the anion is manganate (VII) ion or also known as permanganate ion.

  

c.

The charge of the anion Mno is -1 as evident by the single negative charge on it.

d.

Since the anion has -1 charge, the cation will have +1 charge. Group I elements ( Alkali Metals: Li, Na, K, Rb, Cs, Fr) usually form stable +1 charge cations. Hence, X is likely to be a group I cation.

e.

The mass of MnO4- ion can be calculated as follows:

m(MnO2) = m(Mn)+4x m(O) = 55 g/mol+4 x 16 g/mol = 119 g/mol

Hence, the mass of MnO4 in 3 significant figures is 119 g/mol.

f.

It is given that X accounts for 25% of XMnO4.

Since MnO4 has a mass of 119 g/mol, the mass of X can be calculated as follows:

25 m(X) = 100 x (m(X) + m(MnO4)) m(X) = 0.25(m(X) + 119 g/mol) m(X) - 0.25 x m(X) = 0.25 x 119 g/mol 0.75m(X) = 29.75 g/mol 2

Hence, the mass of X in three significant figures is about 39.7 g/mol.

g.

The group I element with mass closest to 39.7 g/mol is K (Potassium) with a mass of 39.1 g/mol

Hence, the identity of X is K.

h.

Hence, the name of the compound KMnO4 is potassium permanganate.

i.

Formula mass of the compound can be calculated as

m(KMnO4) = m(K)+m(MnO4) = 39.1 g/mol+119 g/mol 158 g/mol

Hence, the formula mass of the compound is 158 g/mol (rounded to 3 significant figures).

j.

Mass of compound taken = 50.5 g

Hence, number of moles can be calculated as

50.5 g moles = mass formula mass 0.320 mol 158 g/mol

k.

Volume of water = 500.0 mL

50.5 g of the compound is dissolved into 500.0 mL water.

Assuming the volume of the solution do not change with addition of the compound, the volume of the solution is 50.5 g.

Hence, there are 0.320 mol of the compound in 500 mL = 0.500 L of solution.

Hence, the concentration in moles/L can be calculated as

0.320 mol -= 0.640 mol/L 0.500 L

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