Chemistry Chapter 10.3 – Flashcards
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A compound is formed when 9.03g Mg combines completely with 3.48g N. What is the % composition of this compound?
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9.03g Mg + 3.48g N= 12.51g--> 9.03/12.51= 72.2% Mg. 3.48/12.51= 27.8% N.
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When a 14.2g sample of murcury(ii) oxide is decomposed into it's elements by heating, 13.2g Hg is obtained. What is the % composition of the compound?
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14.2-13.2= 1--> 13.2/14.2= 92.2% Hg. 1/14.2= 7.1% Murcury(ii) oxide
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Calculate the % composition of these compounds: A. Ethane (C2H6) B. Sodium hydrogen sulfate (NaHSO4)
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A. C2= 24.02 H6= 6--> C2H6= 30.02 24.02/30.02= 80% C. 6/30.02= 20% B. Na= 22.99g. H=1g. O4= 63.96g. S= 32.06--> NaHSO4= 120.01. Na/NaHSO4= 19.2%. H/NaHSO4= 0.81%. S/NaHSO4= 26.7%. O4/NaHSO4= 53.3%
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Calculate the % N in these common fertilizers: A. NH3 B. NH4NO3
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A. N= 14.01 + H3= 3.03-->17.04. N/NH3= 82.2% B. N2= 28.02 + H4= 4.04 + O3= 47.97--> 80.03. N2/NH4NO3= 35%
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Calculate empirical formula of each compound: A. 94.1% O, 5.9% H B. 67.6% Hg, 10.8% S, 21.6% O
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94.1/O(15.99)= 5.88. 5.9/ H(1.01)= 5.85--> 5.88/5.85 1 and 5.85/5.85= 1 so OH= formula B. 67.6/Hg(200.5)= .337. 10.8/S(32.1)= .336. 21.6/O(16)= 1.35--> .337/.336= 1 .336/.336= 1 1.35/.336= 4. Formula= HgSO4
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1, 6-diaminohexane is used to make nylon. what is the empirical formula of this compound if it is 62.1% C, 13.8% H, and 24.1% N?
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62.1/C(12.01)= 5.17. 13.8/H(1.01)= 13.7. 24.1/N(14.01)= 1.7--> 5.17/1.7= 3. 13.7/1.7= 8. 1.7/1.7= 1--> formula= c3h8n
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find the molecular formula of ethylene glycol, which is used as antifreeze. the molar mass is 62g and the empirical formula is CH3O.
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C=12.01 + H3= 3.03 + O=15.99= 31.03. molar mass(62)/31.03= 2--> molecular formula= C2H6O2
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how do you calculate the % by mass of an element in a compound?
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the # of g of the element/ mass in g of the compound, x 100%.
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what information can you obtain from an empirical formula?
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the smallest whole-# ratio of the atoms in a compound.
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how is the molecular formula of a compound related to its empirical formula?
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either the same as its experimentally determined empirical formula, or it is a simple whole-# ratio of the atom in a compound.
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calculate the % composition of the compound that forms when 222.6g N combines with 77.4g O.
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222.6 + 77.4= 300--> 222.6/300= 74.2% N. 77.4/300= 25.8% O
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calculate the % comp. of calcium acitate (Ca(C2H3O2)2).
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Ca=40.08 + C4=48.04 + H6=6.06 + O4=63.96--> 158.14. Ca/158.14=25.3%. C4/158.14=30.4%. H6/158.14=3.7%. O4/158.14=40.4%.
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the compound methyl butanoate smells like apples. its % composition is 58.8% C, 9.8% H, and 31.4 O and its molar mass 102g. Empirical formula? molecular formula?
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empirical formula= 58.8/C(12.01)= 4.9. 9.8/H(1.01)= 9.7. 31.4/O(15.99)= 1.96.--> 4.9/1.96= 2.5. 9.7/1.96= 5. (C2.5H5O)2= c5h10o2 molecular formula= C10H20O4(?)
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Which is an emiprical formula and which is both empirical and molecular?
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a. molecular b. molecular c. both d. both
