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The molecular mass of CBr4 is 12.0 + 4(79.9) = 331.6Amount of CBr4 = mass of substance / molecular mass = 393/331.6 = 1.19mol

This means that a 393g pure sample contains 1.19 moles of tetrabromomethane.

The Avogadro's number is 6.02 x 10^23

So, number of molecules of CBr4 = 1.19 x 6.02 x 10^23 = 7.13 x 10^23

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393 g of CBr4 contain 7,136 572.1023 molecules.

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8y ago
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Q: How many molecules are in 393g of CBr4?
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