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Chapter 3: Problem 84

Determine the molecular formulas to which the following empirical formulas and molar masses pertain. a. \(\operatorname{SNH}(188.35 \mathrm{g} / \mathrm{mol})\) b. \(\mathrm{NPCl}_{2}(347.64 \mathrm{g} / \mathrm{mol})\) c. \(\operatorname{CoC}_{4} \mathrm{O}_{4}(341.94 \mathrm{g} / \mathrm{mol})\) d. \(\mathrm{SN}(184.32 \mathrm{g} / \mathrm{mol})\)

Short Answer

Expert verified
The molecular formulas are: a. \(S_4N_4H_4\) b. \(N_3P_3Cl_6\) c. \(Co_2C_8O_8\) d. \(S_4N_4\)

Step by step solution

01

a. SNH \( (188.35\: g/mol)\)

First, we need to find the molar mass of the empirical formula SNH. The atomic masses of S (sulfur), N (nitrogen) and H (hydrogen) are approximately 32, 14, and 1, respectively. Therefore, the molar mass of the empirical formula SNH is \(32 + 14 + 1 = 47\: g/mol\). Now, we will divide the molecular formula molar mass by the empirical formula molar mass: \(\frac{188.35}{47} \approx 4\) Then, multiply each element in the empirical formula by 4: \(SNH \times 4 = S_4N_4H_4\) The molecular formula is \(S_4N_4H_4\).
02

b. NPCl2 \( (347.64\: g/mol)\)

First, we need to find the molar mass of the empirical formula NPCl2. The atomic masses of N (nitrogen), P (phosphorus), and Cl (chlorine) are approximately 14, 31, and 35.5, respectively. Therefore, the molar mass of the empirical formula NPCl2 is: \(14 + 31 + 2(35.5) = 116\: g/mol\). Now, we will divide the molecular formula molar mass by the empirical formula molar mass: \(\frac{347.64}{116} \approx 3\) Then, multiply each element in the empirical formula by 3: \(NPCl_2 \times 3 = N_3P_3Cl_6\) The molecular formula is \(N_3P_3Cl_6\).
03

c. CoC4O4 \( (341.94\: g/mol)\)

First, we need to find the molar mass of the empirical formula CoC4O4. The atomic masses of Co (cobalt), C (carbon), and O (oxygen) are approximately 59, 12, and 16, respectively. Therefore, the molar mass of the empirical formula CoC4O4 is: \(59 + 4(12) + 4(16) = 171\: g/mol\). Now, we will divide the molecular formula molar mass by the empirical formula molar mass: \(\frac{341.94}{171} \approx 2\) Then, multiply each element in the empirical formula by 2: \(CoC_4O_4 \times 2 = Co_2C_8O_8\) The molecular formula is \(Co_2C_8O_8\).
04

d. SN \( (184.32\: g/mol)\)

First, we need to find the molar mass of the empirical formula SN. The atomic masses of S (sulfur) and N (nitrogen) are approximately 32 and 14, respectively. Therefore, the molar mass of the empirical formula SN is: \(32 + 14 = 46\: g/mol\). Now, we will divide the molecular formula molar mass by the empirical formula molar mass: \(\frac{184.32}{46} \approx 4\) Then, multiply each element in the empirical formula by 4: \(SN \times 4 = S_4N_4\) The molecular formula is \(S_4N_4\).

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Most popular questions from this chapter

Aspartame is an artificial sweetener that is 160 times sweeter than sucrose (table sugar) when dissolved in water. It is marketed as NutraSweet. The molecular formula of aspartame is $\mathrm{C}_{14} \mathrm{H}_{18} \mathrm{N}_{2} \mathrm{O}_{5}$ a. Calculate the molar mass of aspartame. b. What amount (moles) of molecules are present in 10.0 \(\mathrm{g}\) aspartame? c. Calculate the mass in grams of 1.56 mole of aspartame. d. What number of molecules are in 5.0 mg aspartame? e. What number of atoms of nitrogen are in 1.2 g aspartame? f. What is the mass in grams of \(1.0 \times 10^{9}\) molecules of aspartame? g. What is the mass in grams of one molecule of aspartame?

Considering your answer to Exercise \(81,\) which type of formula, empirical or molecular, can be obtained from elemental analysis that gives percent composition?

When aluminum metal is heated with an element from Group 6 \(\mathrm{A}\) of the periodic table, an ionic compound forms. When the experiment is performed with an unknown Group 6 \(\mathrm{A}\) element, the product is 18.56\(\%\) Al by mass. What is the formula of the compound?

Consider the following unbalanced reaction: $$ \mathrm{P}_{4}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{PF}_{3}(g) $$ What mass of \(\mathrm{F}_{2}\) is needed to produce \(120 . \mathrm{g}\) of \(\mathrm{PF}_{3}\) if the reaction has a 78.1\(\%\) yield?

Coke is an impure form of carbon that is often used in the industrial production of metals from their oxides. If a sample of coke is 95\(\%\) carbon by mass, determine the mass of coke needed to react completely with 1.0 ton of copper(Il) oxide. $$ 2 \mathrm{CuO}(s)+\mathrm{C}(s) \longrightarrow 2 \mathrm{Cu}(s)+\mathrm{CO}_{2}(g) $$
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