F. 13 Empirical & Molecular Formula

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Kelly Cai 4D
Posts: 55
Joined: Sat Jul 20, 2019 12:17 am

F. 13 Empirical & Molecular Formula

Postby Kelly Cai 4D » Thu Oct 03, 2019 6:11 pm

In an experiment, 4.14g of phosphorus combined with chlorine to produce 27.8 g of a white solid compound.
A.) What is the empirical formula of the compound?
B.) Assuming that the empirical and molecular formulas of the compound are the same, what is its name?

Tracy Tolentino_2E
Posts: 140
Joined: Sat Sep 07, 2019 12:17 am

Re: F. 13 Empirical & Molecular Formula

Postby Tracy Tolentino_2E » Thu Oct 03, 2019 6:31 pm

So to do this problem, you would need to find the mass percent of phosphorus. Given that we have the grams of phosphorus and the grams of the compound, we can do (4.14 g / 27.8 g) x 100 to get the mass percent, 14.9%. In 100 g of the compound, P would be 14.9 g and we can find Cl to be 85.1 g. To find moles, we divide grams by molar mass, P: 14.9 g/30.973 g per mol = 0.481 moles
Cl: 85.1 g/ 35.453 g per mol = 2.40 moles
Dividing by the lowest moles gives a ratio of 1:5, so the empirical formula is PCl5.

For part b, it's phosphorus pentachloride because they're nonmetals and you name them by the number of atoms.

Camille 4I
Posts: 57
Joined: Sat Aug 24, 2019 12:18 am

Re: F. 13 Empirical & Molecular Formula

Postby Camille 4I » Sun Oct 06, 2019 11:49 pm

For molecules that do not contain metals, do you always determine the name by number of atoms?


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