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Look at Chapter 3 Basic I Homework (New) for the updated questions!
1 mol = x g / MM CH3NH2
MM of CH3NH2 = 12 + 5(1.008) + 14 = 31
1mol= x g / 31 MM
x= 31 g
What is the percent composition by mass of oxygen in carbon dioxide?
carbon dioxide = CO2
C= 12 MM
O =16 MM * 2 = 32
32/ (32+12) = .7272 * 100 = 72.72%
12/ (13.882 +12 +48) = 16.22
What is the mass of 41 molecules of nitrogen dioxide, NO2, in g? USE 3 SIG. FIG.
(14 +16(2)) MM NO2
41/(6.022*10^23) = 6.81e-23 mol = X g/ 46 MM
3.66 mol = X g/ MM of NaCl
MM of NaCl = 58.4428
3.66 mol = X g/ 58.4428 MM
X = 213.9
0 .13 mol = X g/ MM of Na3PO4
MM of Na3PO4 = 163.94
0.13 mol = X g/ 163.94 MM
X = 21.3122
390 g/ MM of CaCO3 = X moles
MM of CaCO3 = 100.0869
390/ 100.0869 = 3.9
X = 3.9 moles
126 g Li2CO3/ 73.891 MM
1.7 moles Li2CO3
The empirical formula indicates the simplest whole number ratio of atoms in a substance.
The molecular formula indicates the number of atoms of each element in a molecule of the substance.
Molecular formula: H2O2; empirical formula: HO
Molecular formulas: C2H4, C5H10; empirical formula: CH2
Molecular formula: CH4; empirical formula: CH4
C2H4O2 / 2
5 g C
5 g / 12 MM = .4167 mol C
1.70 g H / 1.008 MM = 1.686 mol H
1.686 mol H / .4167 = 4.05 about 4
Use a 100g sample...
82.64 g C / 12 MM = 6.89
17.63 g H/ 1.008 MM = 17.49
17.49 / 6.89 = 2.5 (not a whole number) multiply it by 2 to make it a whole number
Left side v Right Side
simply add a 4 coefficient in front of the H2O on the right
C3H8(g) + O2(g) --> CO2(g)+ 4H2O(g)
Next the Cs...
add a 3 coefficient in front of the CO2 on the right
C3H8(g) + O2(g) --> 3CO2(g)+ 4H2O(g)
Next the Os...
add a 5 coefficient in front of the O2 on the left
C3H8(g) + 5O2(g) --> 3CO2(g)+ 4H2O(g)
Now lets recheck to make sure
3 C: L and R
8 H: L and R
10 O: L and R
Balancing equations heck friggin yeah...
Basically all you need to do is get the same amount of stuff on both sides
On the reactants side aka the left side we have:
1 mol of Cu
1 mol of O
1 mol of N
3 mols of H
On the products side aka the right side we have:
1 mol of Cu
1 mol of O
2 mols of N
2 mols of H
... so were going to balance out the N first
and put a coefficient of 2 in front NH3 (on the reactant side) so we can have 2 mols of N on both sides
Nitrogen molecules refer to N2 (btw)
Look at solutions of questions 15