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Alisiya [41]
3 years ago
9

What is the empirical formula for a compound if 300.00 g of it is known to contain 82.46224 g of molybdenum, 45.741 g of chlorin

e and the rest is bromine
Chemistry
1 answer:
Marysya12 [62]3 years ago
8 0

Answer:

MoClBr₂

Explanation:

First we calculate the mass of bromine in the compound:

  • 300.00 g - (82.46224 g + 45.741 g) = 171.79676 g

Then we<u> calculate the number of moles of each element</u>, using their <em>respective molar masses</em>:

  • 82.46224 g Mo ÷ 95.95 g/mol = 0.9594 mol Mo
  • 45.741 g Cl ÷ 35.45 g/mol = 1.290 mol Cl
  • 171.79676 g Br ÷79.9 g/mol = 2.150 mol Br

Now we <u>divide those numbers of moles by the lowest number among them</u>:

  • 0.9594 mol Mo / 0.9594 = 1
  • 1.290 mol Cl / 0.9594 = 1.34 ≅ 1
  • 2.150 mol Br / 0.9594 = 2.24 ≅ 2

Meaning the empirical formula is MoClBr₂.

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