A compound contains 2.8% Fe by mass.If the compound contains single Fe atom in its one molecule, then the molar mass(g/ml) of the compound is? (Fe=56 amu)

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A compound contains 2.8% Fe by mass.If the compound contains single Fe atom in its one molecule, then the molar mass(g/ml) of the compound is? (Fe=56 amu)

The answer is 2000.

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Answer 1 · 2018-03-06T04:17:14

You get 2,000 by basically dividing then multiplying.

Explanation:

If iron, with 56 amu, therefore 56 grams/mol, composes 2.8% of the mass, then for every 2.8% of mass, 56 grams/mol is added to the molar mass.

$\frac{100}{2.8} = 35.714 ~$ $100/2.8 = 35.714~$

That means about 35.714 2.8 percents in the whole compound, so multiply 56 by 35.714~.

$56 \cdot 35.714 ~ = 2000$ $56 * 35.714~ = 2000$

Therefore, molar mass of 2,000 grams/mol.

Answer 2 · 2018-03-06T12:59:39

The molar mass of the compound is 2000 g/mol.

Explanation:

Let's do this by the unitary method.

$2.8 % of the molecular mass = 56 u$ $"2.8 % of the molecular mass = 56 u"$

$1 % of the molecular mass = \frac{56}{2.8} l u$ $"1 % of the molecular mass" = 56/2.8color(white)(l) "u"$

$100 % of the molecular mass = 100 \times \frac{56}{2.8} l u = 2.0 \times 10^{3} l u or 2000 u$ $"100 % of the molecular mass" = 100 × 56/2.8 color(white)(l)"u" = 2.0 × 10^3 color(white)(l)"u or 2000 u"$

If the mass of one molecule is 2000 u, the mass of one mole is 2000 g.

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A compound contains 2.8% Fe by mass.If the compound contains single Fe atom in its one molecule, then the molar mass(g/ml) of the compound is? (Fe=56 amu)

The answer is 2000.

2 Answers

Explanation:

Explanation:

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