Chemistry Chapter 3 Test Questions – Flashcards

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20.9 g ZnI2
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Zinc iodide, ZnI2, can be prepared by the direct combination of elements (Figure 3.3). A chemist determines from the amounts of elements that 0.0654 mol ZnI2 can form. How many grams of zinc iodide is this?
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0.141 mol PbCrO4
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Lead(II) chromate, PbCrO4, is a yellow paint pigment (called chrome yellow) prepared by a precipitation reaction (Figure 3.4). In a preparation, 45.6 g of lead(II) chromate is obtained as a precipitate. How many moles of PbCrO4 is this?
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5.71 x10^22 HCl molecules
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How many molecules are there in a 3.46-g sample of hydrogen chloride, HCl?
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C 40 percent H 6.73 percent O 53.3 percent
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Formaldehyde, CH2O, is a toxic gas with a pungent odor. Large quantities are consumed in the manufacture of plastics (Figure 3.5), and a water solution of the compound is used to preserve biological specimens. Calculate the mass percentages of the elements in formaldehyde (give answers to three significant figures).
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33.4 g
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How many grams of carbon are there in 83.5 g of formaldehyde, CH2O? Use the percentage composition obtained in the previous example (40.0% C, 6.73% H, 53.3% O).
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the empirical formula is Na2Cr2O7
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Chromium forms compounds of various colors. (The word chromium comes from the Greek khroma, meaning "color"; see Figure 3.9.) Sodium dichromate is the most important commercial chromium compound, from which many other chromium compounds are manufactured. It is a bright orange, crystalline substance. An analysis of sodium dichromate gives the following mass percentages: 17.5% Na, 39.7% Cr, and 42.8% O. What is the empirical formula of this compound? (Sodium dichromate is ionic, so it has no molecular formula.)
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C2H4O2
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In Example 3.9, we found the percentage composition of acetic acid to be 39.9% C, 6.7% H, and 53.4% O. Determine the empirical formula. The molecular mass of acetic acid was determined by experiment to be 60.0 amu. What is its molecular formula?
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698 g Fe
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Hematite, Fe2O3, is an important ore of iron; see Figure 3.12. (An ore is a natural substance from which the metal can be profitably obtained.) The free metal is obtained by reacting hematite with carbon monoxide, CO, in a blast furnace. Carbon monoxide is formed in the furnace by partial combustion of carbon. The reaction is Fe2O3(s) + 3CO(g) ---> 2Fe(s) + 3CO2(g) How many grams of iron can be produced from 1.00 kg Fe2O3?
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8.40 g HCl
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Today chlorine is prepared from sodium chloride by electrochemical decomposition. Formerly chlorine was produced by heating hydrochloric acid with pyrolusite (manganese dioxide, MnO2), a common manganese ore. Small amounts of chlorine may be prepared in the laboratory by the same reaction (see Figure 3.13): 4HCl(aq) + MnO2(s) ----> 2H2O(l) + MnCl2(aq) + Cl2(g) How many grams of HCl react with 5.00 g of manganese dioxide, according to this equation?
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0.26 mol
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Zinc metal reacts with hydrochloric acid by the following reaction: Zn(s) + 2HCl(aq) ----> ZnCl2(aq) + H2(g) If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced?
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2.7 g O2
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In a process for producing acetic acid, oxygen gas is bubbled into acetaldehyde, CH3CHO, containing manganese(II) acetate (catalyst) under pressure at 60C. 2CH3CHO(l) + O2(g) ----> 2HC2H3O2(l) In a laboratory test of this reaction, 20.0 g CH3CHO and 10.0 g O2 were put into a reaction vessel. a. How many grams of acetic acid can be produced by this reaction from these amounts of reactants? b. How many grams of the excess reactant remain after the reaction is complete?
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Solutions: 1. In sodium fluoride, there is one atom of sodium and one atom of fluorine. The molar mass will then be: (1 atom x 23 grams/mole of sodium) + (1 atom x 19 grams/mole of fluorine) = 42 grams/mole of sodium fluoride 2. In potassium hydroxide, there is one atom of potassium, one atom of hydrogen, and one atom of oxygen. The molar mass will then be (1 x 39 grams) + (1 x 1 gram) + (1 x 16 grams) = 56 grams/mole of potassium hydroxide 3. In copper (I) chloride, there is one atom of copper and one atom of chlorine. The molar mass is then (1 x 63.5 grams) + (1 x 35.5 grams) = 99 grams/mole of copper (I) chloride 4. In manganese (IV) oxide, there is one atom of manganese and two atoms of oxygen. The molar mass is then (1 x 55 grams) + (2 x 16 grams) = 87 grams/mole of manganese (IV) oxide 5. In calcium sulfate, there is one atom of calcium, one atom of sulfur, and four atoms of oxygen. The molar mass is then (1 x 40 grams) + (1 x 32 grams) + (4 x 16 grams) = 136 grams/mole of calcium sulfate 6. In magnesium phosphate, there are three atoms of magnesium, two atoms of phosphorus, and eight atoms of oxygen. (The formula is Mg3(PO4)2). The molar mass is then (3 x 24 grams) + (2 x 31 grams) + (8 x 16 grams) = 262 grams/mole of magnesium phosphate
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Give the molar masses of the following compounds: 1. sodium fluoride 2. potassium hydroxide 3. copper (I) chloride 4. manganese (IV) oxide 5. calcium sulfate 6. magnesium phosphate
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molecular mass
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the sum of the masses of all atoms in a molecule
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formula mass
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The sum of the atomic masses of all the atoms in a chemical formula
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mole
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the SI base unit used to measure the amount of a substance whose number of particles is the same as the number of atoms of carbon in exactly 12 g of carbon-12
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Avogadro's number
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6.022 x 10^23. The number of particles in exactly one mole of a pure substance
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molar mass
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the mass in grams of one mole of a substance
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percentage composition
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The mass percentages of each element in a compound
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mass percentage
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(mass of solute/mass of solution) x 100%
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empirical (simplest) formula
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chemical formula that shows the kinds of atoms and their relative numbers in a substance in the smallest possible whole-nimber ratios
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stoichiometry
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unit conversion process involving mole equalities.
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limiting reactant
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the reactant that runs out first
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theoretical yield
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the maximum amount of product that could be formed from given amounts of reactants
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percentage yield
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yield/theoretical yield x 100
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a. Formula weight of CH3OH = AW of C + 4(AW of H) + AW of O. Using the values of atomic weights in the periodic table (inside front cover) rounded to four significant figures and rounding the answer to three significant figures, we have FW = 12.01 amu + (4 x 1.008 amu) + 16.00 amu = 32.042 = 32.0 amu b. FW of NO3 = AW of N + 3(AW of O) = 14.01 amu + (3 x 16.00 amu) = 62.01 = 62.0 amu c. FW of K2CO3 = 2(AW of K) + AW of C + 3(AW of O) = (2 x 39.10 amu) + 12.01 amu + (3 x 16.00 amu) = 138.210 = 138 amu d. FW of Ni3(PO4)2 = 3(AW of Ni) + 2(AW of P) + 8(AW of O) = (3 x 58.70 amu) + (2 x 30.97 amu) + (8 x 16.00 amu) = 366.040 = 366 amu
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Formula Masses and Mole Calculations Find the formula masses of the following substances to three significant figures. a. methanol, CH3OH b. nitrogen trioxide, NO3 c. potassium carbonate, K2CO3 d. nickel phosphate, Ni3(PO4)2
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First, find the formula weight of NH4NO3 by adding the respective atomic weights. Then convert it to the molar mass: FW of NH4NO3 = 2(AW of N) + 4(AW of H) + 3(AW of O) = (2 x 14.01 amu) + (4 x 1.008 amu) + (3 x 16.00 amu) = 80.052 amu The molar mass of NH4NO3 = 80.05 g/mol.
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Formula Masses and Mole Calculations Ammonium nitrate, NH4NO3, is used as a nitrogen fertilizer and in explosives. What is the molar mass of NH4NO3?
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5.81 x 10^-3 mol CaSO4 1.16 x 10^-2 mol H2O
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Formula Masses and Mole Calculations Calcium sulfate, CaSO4, is a white, crystalline powder. Gypsum is a mineral, or natural substance, that is a hydrate of calcium sulfate. A 1.000-g sample of gypsum contains 0.791 g CaSO4. How many moles of CaSO4 are there in this sample? Assuming that the rest of the sample is water, how many moles of H2O are there in the sample? Show that the result is consistent with the formula CaSO42H2O.
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Mass percentage carbon = ( sample of mass/sample in C of mass) x 100 = 86.27%
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Mass Percentage A 1.836-g sample of coal contains 1.584 g C. Calculate the mass percentage of C in the coal.
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.14 x 4.15 kg= 0.581 kg N
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Mass Percentage A fertilizer is advertised as containing 14.0% nitrogen (by mass). How much nitrogen is there in 4.15 kg of fertilizer?
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In each part, the numerator consists of the mass of the element in one mole of the compound; the denominator is the mass of one mole of the compound. Use the atomic weights of C = 12.01 g/mol; O = 16.00 g/mol; Na = 22.99 g/mol; H = 1.008 g/mol; P = 30.97 g/mol; Co = 58.93 g/mol; and N = 14.01 g/mol. Percent C = (C of mass/CO of mass) x = 42.9% Percent O = 100.000% - 42.878%C = 57.122 = 57.1%
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Mass Percentage Calculate the percentage composition for each of the following compounds (three significant figures). a. CO
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ethanol
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Mass Percentage Which contains more carbon, 6.01 g of glucose, c6h12o6, or 5.85 g of ethanol, C2H6O?
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Start by calculating the moles of Os and O; then divide each by the smaller number of moles to obtain integers for the empirical formula. Mol Os = 2.16 g Os x 1 mol Os/190.2 g Os =0.01136 mol (smaller number) Mol O =0.0456 mol Integer for Os = 0.01136 ÷ 0.01136 = 1.000 Integer for O = 0.0456 ÷ 0.01136 = 4.01 Within experimental error, the empirical formula is OsO4
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Mass Percentage An oxide of osmium (symbol Os) is a pale yellow solid. If 2.89 g of the compound contains 2.16 g of osmium, what is its empirical formula?
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Mol K =1.013 mol÷ 0.5078 =2 Mol Mn =0.5078 mol (smallest number)÷ 0.5078 =1 Mol O =2.031 mol÷ 0.5078 =4 K2MnO4.
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Mass Percentage Potassium manganate is a dark green, crystalline substance whose composition is 39.6% K, 27.9% Mn, and 32.5% O, by mass. What is its empirical formula?
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The formula weight corresponding to the empirical formula c2h6n may be found by adding the respective atomic weights. Formula weight = 44.08 amu 88.5 amu ÷ 44.1 amu = 2.006, or 2 Therefore, the molecular formula is (C2H6N)2, or c4h12n2.
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Mass Percentage Putrescine, a substance produced by decaying animals, has the empirical formula C2H6N. Several determinations of molecular mass give values in the range of 87 to 90 amu. Find the molecular formula of putrescine
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By inspecting the balanced equation, obtain a conversion factor of eight mol CO2 to two mol C4H10. Multiply the given amount of 0.30 moles of C4H10 by the conversion factor to obtain the moles of H2O. 1.2 mol CO2
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Stoichiometry: Quantitative Relations in Reactions Butane, C4H10, burns with the oxygen in air to give carbon dioxide and water. 2C4H10(g) + 13O2(g) --> 8CO2(g) + 10H2O(g) What is the amount (in moles) of carbon dioxide produced from 0.30 mol C4H10?
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...
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Limiting Reactant; Theoretical and Percentage Yields Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l ) ---> 4KOH(s) + 3O2(g) If a reaction vessel contains 0.25 mol KO2 and 0.15 mol H2O, what is the limiting reactant? How many moles of oxygen can be produced?
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ionic compound
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a compound composed of positive and negative ions
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ionic bond
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a chemical bond resulting from the attraction between oppositely charged ions.
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covalent bond
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bond formed by the sharing of electrons between atoms
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empirical formula
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a formula showing the lowest whole number ratio of atoms in a compound
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structural formula
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A type of molecular notation in which the constituent atoms are joined by lines representing covalent bonds.
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atomic element
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Those elements that exist in nature with single atoms as their basic units.
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hydrate
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a compound that has a specific number of water molecules bound to each formula unit
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mass percent
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the percent by mass of a component of a mixture or of a given element in a compound.
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combustion analysis
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A method used to determine the composition of compounds containing the elements carbon, hydrogen, and occasionally nitrogen.
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molecular formula
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what kind of formula
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uses balls to represent atoms and sticks to represent chemical bonds
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what kind of formula
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space filling model
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a model of a molecule showing the relative sizes of the atoms and their relative orientations
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1. element 2. molecule 3. molecular 4. ionic 5. formula unit
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1-5
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calcium sulfate hemihydrate
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CaSO4•½H2O ; plaster of Paris
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hydrated ionic compounds
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ionic compounds containing a precise number of water molecules. use Greek prefixes for the number of molecules
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balanced
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__CaO + __H?O --> __Ca(OH)?
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2, 1, 2
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__K + __I? --> __KI
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balanced
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__MgO + __SO? --> __MgSO?
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balanced
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__C + __O? --> __CO?
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1, 3, 2, 3
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__Al?(SO?)? + __Ca(OH)? --> __Al(OH)? + __CaSO?
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2, 1, 2, 1
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__Na + __MgF? --> __NaF + __Mg
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3, 2, 2, 3
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__Al + __Cu(NO?)? --> __Cu + __Al(NO?)?
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balanced
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__Mg + __Cl? --> __MgCl?
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1, 2, 1, 1
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__Sn + __HF --> __SnF? + __H?
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balanced
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__C + __O? --> __CO?
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1, 1, 2
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__N? + __O? --> __NO
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2, 1, 2
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__H? + __O? -->__H?O
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