Given the following mass percentages of the elements in...

Given the following mass percentages of the elements in certain compounds, determine the empirical formula in each case and name the compound. (a) $71.89 \% \mathrm{~T} 1 \quad 28.11 \% \mathrm{Br}$ (b) $74.51 \% \mathrm{~Pb} \quad 25.49 \% \mathrm{Cl}$ (c) $82.24 \% \mathrm{~N} \quad 17.76 \% \mathrm{H}$ (d) $72.24 \% \mathrm{Mg} \quad 27.76 \% \mathrm{~N}$

Given the following mass percentages of the elements in certain compounds, determine the empirical formula in each case and name the compound.
(a) $71.89 \% \mathrm{~T} 1 \quad 28.11 \% \mathrm{Br}$
(b) $74.51 \% \mathrm{~Pb} \quad 25.49 \% \mathrm{Cl}$
(c) $82.24 \% \mathrm{~N} \quad 17.76 \% \mathrm{H}$
(d) $72.24 \% \mathrm{Mg} \quad 27.76 \% \mathrm{~N}$

Key Concepts

Chemical Nomenclature

Chemical nomenclature is the systematic method of naming chemical compounds based on their composition and structure. It is important for unambiguously communicating the identity of substances in the scientific community and includes guidelines for naming ionic and covalent compounds.

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms in a compound. It is derived by calculating the mole ratio of elements and simplifying it, ensuring that the relative proportions are maintained without regard to the actual number of atoms in the molecular formula.

Percent Composition Analysis

This concept involves examining the mass percentages of elements in a compound. It allows one to understand the relative amounts of each element present in a sample, which is the first step in determining the compound's empirical formula.

Mole Calculation

Mole calculation is the process of converting the mass of each element to moles using its atomic mass. This conversion is essential because chemical formulas are based on ratios of atoms, and using moles enables a comparison on a consistent scale.

Transcript

00:02 To determine the empirical formula of each of these compounds from the percent mass, we simply need to assume we have 100 grams of the compound.

00:11 If we have 100 grams of the compound, then at 71 .89 % thallium, we have 71 .89 grams of thallium, which we can divide by the molar mass of thalium to get moles of thalium in the compound.

00:25 0 .352.

00:28 If we have 28 .11 % bromine, 100 grams of the compound has 28.

00:33 0 .11 grams of bromine.

00:35 Dividing that by the molar mass, 79 .904, gives us 0 .352 moles of bromine in the compound.

00:44 So with 0 .352 moles of bromine, a thalium and 0 .352 moles of bromine in the compound, we have equal moles of both of them.

00:54 Dividing both by one, we get 1 mole thalium for every 1 mole of bromine.

00:58 So the empirical formula is t -l -b -r, and the name is thallium bromide.

01:04 We may want to say thallium one bromide.

01:08 As the charge is one and thallium is a metal.

01:11 It's not a transition metal.

01:13 So commonly the roman numeral is not used, but neither is lead, and a roman numeral is used with lead.

01:19 So it's correct to say thallium one bromide.

01:24 For the next compound, it's 74 .51 % lead.

01:29 So if we have 100 grams of the compound, we have 74 .51 grams of lead, which we can divide by the molar mass of lead to get the moles of lead in the compound.

01:40 0 .3596.

01:43 If that same compound is 25 .49 % chlorine, then 100 grams of the compound has 25 .49 grams of chlorine, which we can divide by the molar mass in order to get the moles of chlorine in the compound.

01:56 So we have 0 .3596 moles of lead for every 0 .7 -190 moles of chlorine...

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