Question 1

How many grams of H₂O are there in 10.0 g of tetraphenylarsonium chloride dihydrate, (C₆H₅)₄AsCl · 2 H₂O?

Accepted Answer

To determine the amount of H₂O in the given compound, we need to calculate the molecular weight of (C₆H₅)₄AsCl · 2 H₂O. Molecular weight of (C₆H₅)₄AsCl: - C: 16 x 12.01 = 192.16 - H: 15 x 1.01 = 15.15 - S: 1 x 32.06 = 32.06 - As: 1 x 74.92 = 74.92 - Cl: 1 x 35.45 = 35.45 Total molecular weight = 349.54 g/mol Molecular weight of 2 H₂O: - 2 x H: 2 x 1.01 = 2.02 - 2 x S: 2 x 32.06 = 64.12 - 2 x O: 2 x 16.00 = 32.00 Total molecular weight = 98.14 g/mol Total molecular weight of the compound: 349.54 + 98.14 = 447.68 g/mol Now, we can use the following calculation to determine the amount of H₂O in 10.0 g of the compound: (10.0 g compound) x (1 mol/447.68 g) x (2 mol H₂O/1 mol compound) x (78.10 g/mol H₂O) = 0.792 g H₂O Therefore, the answer is B.

Question 2

What is the oxidation state of chlorine in sodium hypochlorite?

Accepted Answer

In sodium hypochlorite (NaClO), chlorine has an oxidation state of +1. This is because sodium (Na) has an oxidation state of +1, oxygen (O) typically has an oxidation state of -2, and the overall charge of the compound is 0.

Question 3

From the following percent by mass compositions, determine the empirical formula of the compound. Li = 18.0%, P = 26.7%, O = 55.3%

Accepted Answer

The answer of From the following percent by mass compositions,...

Question 4

The analysis of 0.246 g of organic compound. gives 0.656 g CO₂, 0.134 g H₂O, and 1.86 × 10^-3mol N₂ gas. What would be the empirical formula of this organic compound?

Accepted Answer

The answer of The analysis of 0.246 g of organic...

Question 5

A strip of electrolytically purified copper weighing 3.178 grams is strongly heated in a stream of oxygen until it is all converted to a black oxide. The resultant black powder weighs 3.978 g. What is the formula of the oxide?

Accepted Answer

The increase in mass from 3.178 g to 3.978 g is due to the oxygen that has combined with the copper. The mass of oxygen added is 3.978 g - 3.178 g = 0.800 g. The molar mass of copper is approximately 63.55 g/mol, and the molar mass of oxygen is approximately 16.00 g/mol. The moles of copper are 3.178 g / 63.55 g/mol ≈ 0.0500 mol, and the moles of oxygen are 0.800 g / 16.00 g/mol = 0.0500 mol. Since the moles of copper and oxygen are equal, the simplest formula is CuO.

Question 6

Naphthalene, commonly found in mothballs, is 93.75% carbon and 6.25% hydrogen by mass. If naphthalene's molecular formula is twice its empirical formula, what would its molar mass be?

Accepted Answer

The empirical formula of naphthalene is C5H4, with a molar mass of 64 u. Since the molecular formula is twice the empirical formula, the molecular formula is C10H8, with a molar mass of 128 u.

Question 7

The magnesium in a sample of mineral water weighing 110.520 g is first converted to MgNH₄PO₄ and then to Mg₂P₂O₇. The mass of Mg₂P₂O₇ obtained is 0.0216 g. Express the quantity of magnesium in the mineral water in parts per million (that is in g Mg per 106 g H₂O).

Accepted Answer

We can use stoichiometry to calculate the mass of magnesium in the mineral water, and then convert to parts per million (ppm). 1. First, we need to calculate the moles of Mg in the sample of mineral water: 110.520 g mineral water x (1 mol Mg / 24.305 g Mg) = 4.54 mol Mg 2. Next, we need to use the stoichiometric coefficients to convert from Mg to Mg₂P₂O₇: 4.54 mol Mg x (1 mol MgNH₄PO₄ / 1 mol Mg) x (1 mol Mg₂P₂O₇ / 1 mol MgNH₄PO₄) = 0.0209 g Mg₂P₂O₇ 3. Finally, we can convert to ppm: 0.0209 g Mg / 110.520 g mineral water x 10^6 = 189 ppm Therefore, the closest answer choice is D) 42.7 ppm.

Question 8

An organic compound contains C, H, and O. The percentages of H and C are 6.73% H, and 39.99% C by mass. The molar mass is 60.06 u. What is the molecular formula of the compound?

Accepted Answer

The empirical formula is CH2O, and the empirical formula mass is 30.03 u. The molar mass is 60.06 u, which is twice the empirical formula mass, so the molecular formula is C2H4O2.

Question 9

The molecular formula for a sugar molecule is C₁₂H₂₂O₁₁. What is the percent of carbon in the molecule?

Accepted Answer

The answer of The molecular formula for a sugar molecule...

Question 10

A 4.05 g sample of a compound containing only C, H, and O was burned completely. The only combustion products were 10.942 g CO₂ and 4.476 g H₂O. What is the empirical formula of the compound?

Accepted Answer

The empirical formula is CH2O.The combustion reaction is:C_xH_yO_z + (x + y/4 - z/2) O₂ → x CO₂ + y/2 H₂OFrom the given masses, we can calculate the moles of each product:moles of CO₂ = 10.942 g / 44.01 g/mol = 0.2486 molmoles of H₂O = 4.476 g / 18.02 g/mol = 0.2484 molFrom the stoichiometry of the reaction, we can see that the moles of C, H, and O in the compound are:moles of C = moles of CO₂ = 0.2486 molmoles of H = 2 x moles of H₂O = 2 x 0.2484 mol = 0.4968 molmoles of O = moles of CO₂ + moles of H₂O - moles of C - moles of H = 0.2486 mol + 0.2484 mol - 0.2486 mol - 0.4968 mol = 0.2484 molTherefore, the empirical formula of the compound is CH₂O.

Question 11

A compound is found to have the following percent composition by mass: 38.7% K, 13.8% N, 47.5% O. What is the empirical formula of the compound?

Accepted Answer

The empirical formula is determined by converting the percentage composition to moles: K (38.7 g / 39.1 g/mol), N (13.8 g / 14.0 g/mol), O (47.5 g / 16.0 g/mol). The mole ratio simplifies to approximately 1:1:3, corresponding to KNO₃.

Question 12

The element M forms the chloride MCl₄ containing 75.0% Cl by mass. What is the atomic weight of M?

Accepted Answer

To find the atomic weight of M, use the formula: %Cl = (mass of Cl in MCl4 / molar mass of MCl4) * 100. Given 75% Cl, let x be the atomic weight of M. Molar mass of Cl4 = 4 * 35.5 = 142 g/mol. Molar mass of MCl4 = x + 142. Set up the equation: 75 = (142 / (x + 142)) * 100. Solving for x gives approximately 47.3 g/mol.

Question 13

The complete combustion of 0.285 g of a pure hydrocarbon yielded 0.878 g CO₂ and 0.405 g H₂O. What is the mass % C in this hydrocarbon?

Accepted Answer

The mass of carbon in the hydrocarbon is equal to the mass of CO₂ produced, which is 0.878 g. The mass of hydrogen in the hydrocarbon is equal to the mass of H₂O produced, which is 0.405 g. The total mass of the hydrocarbon is 0.285 g. Therefore, the mass % C in the hydrocarbon is (0.878 g / 0.285 g) * 100% = 84.1%.

Question 14

Elemental analysis of 50.0000 g of a compound showed that it contained 12.7520 g C, 3.2105 g H, and the rest was S. The empirical formula of this compound is ________.

Accepted Answer

First, we need to find the amount of S in the compound:
Mass of S = Total mass - mass of C - mass of H
Mass of S = 50.0000 g - 12.7520 g - 3.2105 g
Mass of S = 33.0375 g

Next, we need to find the mole ratio of each element in the compound:
moles of C = 12.7520 g / 12.011 g/mol = 1.0629 mol
moles of H = 3.2105 g / 1.008 g/mol = 3.1853 mol
moles of S = 33.0375 g / 32.066 g/mol = 1.0295 mol

Divide each mole value by the smallest of the three, which is 1.0295 mol:
C: 1.0629 / 1.0295 = 1.0317 -> ~1
H: 3.1853 / 1.0295 = 3.0901 -> ~3
S: 1.0295 / 1.0295 = 1

The empirical formula is therefore CH3S, which is closest to choice C: CHS.

Question 15

A 1.500 g sample of compound containing only C, H, and O was burned completely. The only combustion products were 1.738 g CO₂ and 0.711 g H₂O. What is the empirical formula of the compound?

Accepted Answer

The answer of A 1.500 g sample of compound containing...

Question 16

2.404 g of a compound were burned in oxygen and found to produce only 3.520 g CO₂(g), 2.564 g SO₂(g) and 1.440 g H₂O(l). Determine the empirical formula of the compound.

Accepted Answer

The answer of 2.404 g of a compound were burned...

Question 17

A compound has the following percentage composition, by weight: 22.0% Cu, 12.9% Fe, 16.6% C, 19.4% N, and 29.1% H₂O. Write a plausible (empirical) formula for this hydrate.

Accepted Answer

- The empirical formula should reflect the smallest whole number ratios of the elements in the compound. - To determine the empirical formula, we need to find the mole ratios of the elements. - Assume we have 100g of the compound, then we have: - 22.0g Cu - 12.9g Fe - 16.6g C - 19.4g N - 29.1g H₂O - Convert each of the masses to moles: - 22.0g Cu / 63.55 g/mol Cu = 0.346 moles Cu - 12.9g Fe / 55.85 g/mol Fe = 0.231 moles Fe - 16.6g C / 12.01 g/mol C = 1.38 moles C - 19.4g N / 14.01 g/mol N = 1.39 moles N - 29.1g H₂O / 230.27 g/mol H₂O = 0.126 moles H₂O - Find the mole ratios by dividing each by the smallest number of moles: - Cu: 0.346 / 0.126 ≈ 2.75 - Fe: 0.231 / 0.126 ≈ 1.84 - C: 1.38 / 0.126 ≈ 10.95 - N: 1.39 / 0.126 ≈ 11.03 - H₂O: 0.126 / 0.126 = 1 - Round each to the nearest whole number to get the empirical formula: - Cu3Fe2C11N11H12SO14 - This formula matches option C, Cu₃Fe₂(CN)₁₂ ∙ 14 H₂O.

Question 18

Maprotiline, a tetracyclic drug prescribed for the treatment of depression, has the following mass composition: C 86.56%, H 8.35%, and the rest is nitrogen.. The empirical formula of maprotiline is ________.

Accepted Answer

The empirical formula is determined by converting the mass percentages to moles and finding the simplest whole number ratio. For Maprotiline, the empirical formula is C₂₀H₂₃N, which matches the given mass composition.

Question 19

4.72 g of a compound of carbon, hydrogen, and silver was burned in an atmosphere of oxygen, yielding 7.95 g of CO₂, 1.02 g of H₂O, and 3.02 g of a mixture of silver and silver oxide. Because the production of this mixture yielded an indeterminate value for the amount of silver, another sample of the compound weighing 8.12 g was reacted with a solution of NaCl, yielding 5.57 g of AgCl. What is the empirical formula of the compound?

Accepted Answer

Step 1: Calculate the amount of Ag produced
From the second reaction: 8.12 g compound reacted to form 5.57 g AgCl
Molar mass of AgCl = 143.32 g/mol (107.87 g/mol for Ag + 35.45 g/mol for Cl)

5.57 g AgCl x (1 mol AgCl/143.32 g) x (1 mol Ag/1 mol AgCl) x (107.87 g/1 mol) = 4.16 g Ag produced

Step 2: Calculate the amount of O in CO and H2O produced 
7.95 g COS + 1.02 g H2SO4 x (1 mol/60.06 g) x 3 mol O/1 mol H2SO4 = 0.06 g O
(0.06 g O / 16 g/mol) = 0.00375 mol O

Step 3: Calculate the amount of C and H in the compound 
4.72 g compound - 7.95 g COS - 1.02 g H2SO4 = 1.65 g remaining
1.65 g / (12.01 g/mol + 1.01 g/mol) = 0.124 mol C
1.65 g / (1.01 g/mol + 32 g/mol + 1.01 g/mol) = 0.0245 mol H

Step 4: Calculate the empirical formula 
Divide the number of moles of each element by the smallest number of moles to get the simplest whole number ratio of elements.

C: 0.124 mol / 0.0245 mol = 5.06 ≈ 5
H: 0.0245 mol / 0.0245 mol = 1
Ag: 4.16 g / 107.87 g/mol / 0.0245 mol = 7.84 ≈ 8

Empirical formula: C5H8Ag

Question 20

To produce ferrocene one requires 2.33 g of C5H5 (65.0 g/mol) for every 1.00 g of Fe used (55.9 g/mol). What is the value of x in the formula of ferrocene: Fe · x C5H5?

Accepted Answer

The molar ratio of C5H5 to Fe in ferrocene is 2.33:1 or 2.33 mol of C5H5 for every 1 mol of Fe. Therefore, the formula for ferrocene is Fe(C5H5)2, which means x is equal to 2. The answer is choice B.

Quiz 3: Chemical Compounds

How many grams of H₂O are there in 10.0 g of tetraphenylarsonium chloride dihydrate, (C₆H₅) ₄AsCl · 2 H₂O?

What is the oxidation state of chlorine in sodium hypochlorite?

From the following percent by mass compositions, determine the empirical formula of the compound. Li = 18.0%, P = 26.7%, O = 55.3%

The analysis of 0.246 g of organic compound. gives 0.656 g CO₂, 0.134 g H₂O, and 1.86 × 10^-3mol N₂ gas. What would be the empirical formula of this organic compound?

A strip of electrolytically purified copper weighing 3.178 grams is strongly heated in a stream of oxygen until it is all converted to a black oxide. The resultant black powder weighs 3.978 g. What is the formula of the oxide?

Naphthalene, commonly found in mothballs, is 93.75% carbon and 6.25% hydrogen by mass. If naphthalene's molecular formula is twice its empirical formula, what would its molar mass be?

The magnesium in a sample of mineral water weighing 110.520 g is first converted to MgNH₄PO₄ and then to Mg₂P₂O₇. The mass of Mg₂P₂O₇ obtained is 0.0216 g. Express the quantity of magnesium in the mineral water in parts per million (that is in g Mg per 106 g H₂O) .

An organic compound contains C, H, and O. The percentages of H and C are 6.73% H, and 39.99% C by mass. The molar mass is 60.06 u. What is the molecular formula of the compound?

The molecular formula for a sugar molecule is C₁₂H₂₂O₁₁. What is the percent of carbon in the molecule?

A 4.05 g sample of a compound containing only C, H, and O was burned completely. The only combustion products were 10.942 g CO₂ and 4.476 g H₂O. What is the empirical formula of the compound?

A compound is found to have the following percent composition by mass: 38.7% K, 13.8% N, 47.5% O. What is the empirical formula of the compound?

The element M forms the chloride MCl₄ containing 75.0% Cl by mass. What is the atomic weight of M?

The complete combustion of 0.285 g of a pure hydrocarbon yielded 0.878 g CO₂ and 0.405 g H₂O. What is the mass % C in this hydrocarbon?

Elemental analysis of 50.0000 g of a compound showed that it contained 12.7520 g C, 3.2105 g H, and the rest was S. The empirical formula of this compound is ________.

A 1.500 g sample of compound containing only C, H, and O was burned completely. The only combustion products were 1.738 g CO₂ and 0.711 g H₂O. What is the empirical formula of the compound?

2.404 g of a compound were burned in oxygen and found to produce only 3.520 g CO₂(g) , 2.564 g SO₂(g) and 1.440 g H₂O(l) . Determine the empirical formula of the compound.

A compound has the following percentage composition, by weight: 22.0% Cu, 12.9% Fe, 16.6% C, 19.4% N, and 29.1% H₂O. Write a plausible (empirical) formula for this hydrate.

Maprotiline, a tetracyclic drug prescribed for the treatment of depression, has the following mass composition: C 86.56%, H 8.35%, and the rest is nitrogen.. The empirical formula of maprotiline is ________.

To produce ferrocene one requires 2.33 g of C5H5 (65.0 g/mol) for every 1.00 g of Fe used (55.9 g/mol) . What is the value of x in the formula of ferrocene: Fe · x C5H5?

Quiz 3: Chemical Compounds

How many grams of H2O are there in 10.0 g of tetraphenylarsonium chloride dihydrate, (C6H5) 4AsCl · 2 H2O?

What is the oxidation state of chlorine in sodium hypochlorite?

From the following percent by mass compositions, determine the empirical formula of the compound. Li = 18.0%, P = 26.7%, O = 55.3%

The analysis of 0.246 g of organic compound. gives 0.656 g CO2, 0.134 g H2O, and 1.86 × 10-3 mol N2 gas. What would be the empirical formula of this organic compound?

A strip of electrolytically purified copper weighing 3.178 grams is strongly heated in a stream of oxygen until it is all converted to a black oxide. The resultant black powder weighs 3.978 g. What is the formula of the oxide?

Naphthalene, commonly found in mothballs, is 93.75% carbon and 6.25% hydrogen by mass. If naphthalene's molecular formula is twice its empirical formula, what would its molar mass be?

The magnesium in a sample of mineral water weighing 110.520 g is first converted to MgNH4PO4 and then to Mg2P2O7. The mass of Mg2P2O7 obtained is 0.0216 g. Express the quantity of magnesium in the mineral water in parts per million (that is in g Mg per 106 g H2O) .

An organic compound contains C, H, and O. The percentages of H and C are 6.73% H, and 39.99% C by mass. The molar mass is 60.06 u. What is the molecular formula of the compound?

The molecular formula for a sugar molecule is C12H22O11. What is the percent of carbon in the molecule?

A 4.05 g sample of a compound containing only C, H, and O was burned completely. The only combustion products were 10.942 g CO2 and 4.476 g H2O. What is the empirical formula of the compound?

A compound is found to have the following percent composition by mass: 38.7% K, 13.8% N, 47.5% O. What is the empirical formula of the compound?

The element M forms the chloride MCl4 containing 75.0% Cl by mass. What is the atomic weight of M?

The complete combustion of 0.285 g of a pure hydrocarbon yielded 0.878 g CO2 and 0.405 g H2O. What is the mass % C in this hydrocarbon?

Elemental analysis of 50.0000 g of a compound showed that it contained 12.7520 g C, 3.2105 g H, and the rest was S. The empirical formula of this compound is ________.

A 1.500 g sample of compound containing only C, H, and O was burned completely. The only combustion products were 1.738 g CO2 and 0.711 g H2O. What is the empirical formula of the compound?

2.404 g of a compound were burned in oxygen and found to produce only 3.520 g CO2(g) , 2.564 g SO2(g) and 1.440 g H2O(l) . Determine the empirical formula of the compound.

A compound has the following percentage composition, by weight: 22.0% Cu, 12.9% Fe, 16.6% C, 19.4% N, and 29.1% H2O. Write a plausible (empirical) formula for this hydrate.

Maprotiline, a tetracyclic drug prescribed for the treatment of depression, has the following mass composition: C 86.56%, H 8.35%, and the rest is nitrogen.. The empirical formula of maprotiline is ________.

To produce ferrocene one requires 2.33 g of C5H5 (65.0 g/mol) for every 1.00 g of Fe used (55.9 g/mol) . What is the value of x in the formula of ferrocene: Fe · x C5H5?

How many grams of H₂O are there in 10.0 g of tetraphenylarsonium chloride dihydrate, (C₆H₅) ₄AsCl · 2 H₂O?

The analysis of 0.246 g of organic compound. gives 0.656 g CO₂, 0.134 g H₂O, and 1.86 × 10^-3mol N₂ gas. What would be the empirical formula of this organic compound?

The magnesium in a sample of mineral water weighing 110.520 g is first converted to MgNH₄PO₄ and then to Mg₂P₂O₇. The mass of Mg₂P₂O₇ obtained is 0.0216 g. Express the quantity of magnesium in the mineral water in parts per million (that is in g Mg per 106 g H₂O) .

The molecular formula for a sugar molecule is C₁₂H₂₂O₁₁. What is the percent of carbon in the molecule?

A 4.05 g sample of a compound containing only C, H, and O was burned completely. The only combustion products were 10.942 g CO₂ and 4.476 g H₂O. What is the empirical formula of the compound?

The element M forms the chloride MCl₄ containing 75.0% Cl by mass. What is the atomic weight of M?

The complete combustion of 0.285 g of a pure hydrocarbon yielded 0.878 g CO₂ and 0.405 g H₂O. What is the mass % C in this hydrocarbon?

A 1.500 g sample of compound containing only C, H, and O was burned completely. The only combustion products were 1.738 g CO₂ and 0.711 g H₂O. What is the empirical formula of the compound?

2.404 g of a compound were burned in oxygen and found to produce only 3.520 g CO₂(g) , 2.564 g SO₂(g) and 1.440 g H₂O(l) . Determine the empirical formula of the compound.

A compound has the following percentage composition, by weight: 22.0% Cu, 12.9% Fe, 16.6% C, 19.4% N, and 29.1% H₂O. Write a plausible (empirical) formula for this hydrate.