Chem 170 chapter 3 Flashcards
How many atoms of Br are in 35.2g of CH2Br2
ANSWER: 2.44x10^23 amu Br
How?
add element masses together of CH2Br2 = 174.0 g/mol
35.2g x (1mol/174gmol) = 0.202
Br(2) x (0.202) = 0.404
(0.404) x (6.022x10^23”avagadros” / 1mol) = 2.44x10^23 amu Br
Determine the chemical formula for this experimental data. Se 1.443g Br 5.841g
(Se 1.443g) / (Se = 78.96g/mol) = 0.018
(Br 5.841g) / (Br = 79.9g/mol) = 0.073
divide by the smallest #
(0.073) / (0.018) = 4.055
Answer = Se1Br4
Calculate the mass of sodium in 8.5g of Na3PO4
Na = 22.990 x (3) = 68.97 P = 30.974 x (1) = 30.974 O = 15.998 x (4) = 63.992 Total = 163.936
8.5 x (1mol / 163.936) = 0.05 mole
Na3PO4 = 0.05 Na : (3) x 0.05 = 0.15
0.15 x (23 / 1moll) = 3.45
Find moles atoms & grams of an element in a compound
0.0884mol C4H10
C: (4) x 0.0884 = 0.3536
H: (10) x 0.0884 = 0.884
Find moles atoms & grams of an element in a compound
2.4+mol C6H12
C: (6) x 2.4 = 14.4
H: (12) x 2.4 = 28.8
What is the emperical formula for C10H8
C5H4
Which substance is an ionic compound and why? SrI2 N2O4 He CCl4
SrI2
What is the correct formula for a compound between calcium and sulfur?
CaS
just be sure its balanced 2+ & 2-
Name the compound SrI2
Stronlium Iodide
What is the formula for manganese(IV) oxide?
MnO2
Name the compound Pb(C2H3O2)2
lead(II) acetate
Name the compound P2I4
diphosphorus tetraiodide
Name the compound HNO2(aq)
nitrous acid
Determine the number of CH2Cl2 molecules in 25g of CH2Cl2
1.77 x 10^23 molecules
.294350 X 10^23 = 1.77 x 10^23 molecules
list the elements in the compound CF2Cl2 in order of decreasing mass percent composition
Cl > F > C
Determine the mass of potassium in 35.5g of KBr
11.7g
- 5 / 119.002 (KBRmass) = 0.298314314
- 298314314 * 39.098 (K) = 11.66348077
A compound is 52.14% C, 13.13% H, and 34.73% O by mass. What is the empirical formula of the compound?
% H = 100 - ( 52.14 + 34.73 )=13.13 %
assume 100 g of this compound
mass H = 13.13 g
moles H = 13.13 g / 1.008 g/mol=13
mass C = 52.14 g
moles C = 52.14 g/ / 12.011 g/mol=4
mass O = 34.73 g
moles O = 34.73 g/ 15.999 g/mol=2
the empirical formula is C4H13O2
A compound has the empirical formula CH2O and a formula mass of 120.10 amu. What is the molecular formula of the compound?
C4H8O4
Combustion of 30.42g of a compound containing only carbon, hydrogen, and oxygen produces 35.21 g of CO2 and 14.42 g of H2O. What is the empirical formula of the compound?
a) C6HO12 b) C2H2O3 c) C2H4O3 d) C4H8O6 e) CHO
C2H4O3
Combustion analysis yields information about compound C and H content only, so we can determine the mole ratio of C:H in the compound by the following: (35.21 g CO 2 )(1 mol CO 2 / 44 g)(1 mol C / mol CO 2 ) = 0.8 mol
C (14.42 g H 2 O)(1 mol H 2 O / 18 g)(2 mol H / 1 mol H 2 O) = 1.6 mol
H 0.8 mol C : 1.6 mol H :: 1 mol C : 2 mol H
Thus, the empirical formula must be a lowest whole number formula giving a 1:2 mole ratio for C:H, which only answer (c) does. Answer (c) is correct.