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Vlada [557]
3 years ago
9

In an experiment, a compound was determined to contain 68.94% oxygen and 31.06% of an unknown element by weight. The molecular w

eight of the compound is 69.7 g/mol. What is this compound?\
Chemistry
1 answer:
monitta3 years ago
8 0

Answer is: the compound is B₂O₃.

ω(O) = 68.94% ÷ 100%.

ω(O) = 0.6894; percentage of oxygen in the compound.

ω(X) = 31.06% ÷ 100%.

ω(X) = 0.3106; percentage of unknown element in the compound.

If we take 69.7 grams of the compound:

M(compound) = 69.7 g/mol.

n(compound) = 69.7 g ÷ 69.7 g/mol.

n(compound) = 1 mol.

n(O) = (69.7 g · 0.6894) ÷ 16 g/mol.

n(O) = 3 mol.

M(compound) = n(O) · M(O) + n(X) · M(X).

n(X) = 1 mol ⇒ M(X) = 21.7 g/mol; there is no element with this molecular weight.

n(X) = 2 mol ⇒ M(X) = 10.85 g/mol; this element is boron (B).

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Ammonium phosphate NH43PO4 is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric a
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<em />

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How many grams of the molecule (C2H4O2) would be required to make 1 L of a 0.5 M solution of the molecule? (Carbon = 12, Oxygen
NeTakaya

MC₂H₄O₂ = 2 * 12 g/mol + 4 * 1 g/mol + 2 * 16 g/mol = 60 g/mol

c = 0.5 mol/L

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n = c * V

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