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mote1985 [20]
3 years ago
11

The mineral enargite is 48.41% cu, 19.02% as, and 32.57% s by mass. what is the empirical formula of enargite?

Chemistry
1 answer:
Firlakuza [10]3 years ago
4 0
Answer is: <span>the empirical formula of enargite is Cu</span>₃AsS₄.<span>
If we use 100 grams of enargite:
n(Cu) = 48,41 g </span>÷ 63.55 g/mol.
n(Cu) = 0.761 mol.
n(As) = 19.02 g ÷ 74.92 g/mol.
n(As) = 0.254 mol.
n(S) = 32.57 g ÷ 32.065 g/mol.
n(S) = 1.016 mol.
n(Cu) : n(As) : n(S) = 0.761 mol : 0.254 mol : 1.016 mol / 0.254.
n(Cu) : n(As) : n(S) = 3 mol : 1 mol : 4 mol.
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When of a certain molecular compound X are dissolved in of benzene , the freezing point of the solution is measured to be . Calc
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The question is incomplete. Here is the complete question.

When 2.10 g of a certain molecular compound X are dissolved in 65.0 g of benzene (C₆H₆), the freezing point of the solution is measured to be 3.5°C. Calculate the molar mass of X. If you need any additional information on benzene, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to 2 significant digits.

Answer: MM = 47.30 g/mol.

Explanation: There is a relationship between <u>freezing</u> <u>point</u> <u>depression</u> and <u>molality</u>. With this last one, is possible to calculate <u>molar</u> <u>mass</u> or molar weight of a compound.

<u>Freezing</u> <u>Point</u> <u>Depression</u> occurs when a solute is added to a solvent: the freezing point of the solvent decreases when a non-volatile solute is incremented.

<u>Molality</u> or <u>molal</u> <u>concentration</u> is a quantity of solute dissolved in a certain mass, in kg, of solvent. Its symbol is m and it's defined as

m=\frac{moles(solute)}{kg(solvent)}

Freezing point depression and molal are related as the following:

\Delta T_{f}=K_{f}.m

where

\Delta T_{f} is freezing point depression of solution

K_{f} is molal freezing point depression constant

m is molality

Now, to determine molar mass, first, find molality of the mixture:

\Delta T_{f}=K_{f}.m

m=\frac{\Delta T_{f}}{K_{f}}

For benzene, constant is 5.12°C/molal. Then

m=\frac{3.5}{5.12}

m = 0.683 molal

Second, knowing the relationship between molal and moles of solute, determine the last one:

m=\frac{moles(solute)}{kg(solvent)}

mol(solute)=m.kg(solvent)

mol(solute) = 0.683(0.065)

mol(solute) = 0.044 mol

The definition for <u>Molar</u> <u>mass</u> is the mass in grams of 1 mol of substance:

n(moles)=\frac{m(g)}{MM(g/mol)}

MM=\frac{m}{n}

In the mixture, there are 0.044 moles of X, so its molecular mass is

MM=\frac{2.1}{0.044}

MM = 47.30 g/mol

The molecular compound X has molecular mass of 47.30 g/mol.

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