What is the empirical and molecular for a compound that is 36.1% Sr and 63.9% Br with a molar mass of 742 g/mo

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What is the empirical and molecular for a compound that is 36.1% Sr and 63.9% Br with a molar mass of 742 g/mo

[From: ] [author: ] [Date: 11-05-25] [Hit: ]

1 g Sr) / (87.6210 g/mol) = 0.(63.9 g Br) / (79.9041 g/mol) = 0.Divide by the smaller number of moles:0.......

Please help with this question.

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Take a hypothetical 100 g sample of the compound.

(36.1 g Sr) / (87.6210 g/mol) = 0.412001689 mole Sr
(63.9 g Br) / (79.9041 g/mol) = 0.799708651 mole Br

Divide by the smaller number of moles:
0.412001689 / 0.412001689 = 1.000
0.799708651 / 0.412001689 = 1.941
Round to integers. So the empirical formula is:
SrBr2

The molecular mass of SrBr2 is 247.4292 g/mol.
(742 g/mo) / (247.4292 g/mol) = 2.99884
Round to the nearest integer, 3.
Multiply the empirical formula by 3 to find the molecular formula:
Sr3Br6

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