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Chemistry Review (12/8/11]  Moles WKSHT
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Chemistry Review (12/8/11]  Moles WKSHT
This is really, REALLY simple to me, but you probably aren't me so I'll attempt to walk it through. Sorry for any confusion, of course if you have any questions or concerns you can always PM me or post in this topic for clarification!
1) Define "mole".
Mole is a unit of measure for a substance.
2) How many moles are present in 34 grams of Cu(OH)_{2}?
To find out how many moles are present in X amount of anything, remember that Moles x Formula Mass equals the grams. So, find out the mass of Cu(OH)_{2} which isCu: 63.543
O_{2}: 15.999*2 : 31.998
H_{2}: 2.106
Which equals 97.647. So we have the formula mass (I'll call it F) and we have the given mass (34, which I'll call G). Moles I'll call M.
M x F = G
97.647M = 34
Working backwards, we say 34/97.647 = about .35 moles.
3) How many moles are present in 2.45x10^{23} molecules of CH_{4}?
Recall fromMole Potpourri and other sources that
P/A = Mol
Mol x m_{base formula} = m_{sample amount}
Or in other words, The particle amount divided by A's # = the moles AND
Moles times mass of base formula = mass of the sample amount.
Luckily, we're only trying to get to the moles, so discard the second equation [for now].
We know the particle amount of CH_{4} sample is 2.45x10^{23}. And we also know A's number is 6.02x10^{23}. Divide 2.45x10^{23} by 6.02x10^{23} and you'll get approximately .41 moles.
4) How many grams are there in 3.4x10^{24} molecules of NH_{3}?
Recall fromMole Potpourri and other sources that
P/A = Mol
Mol x m_{base formula} = m_{sample amount}
Or in other words, The particle amount divided by A's # = the moles AND
Moles times mass of base formula = mass of the sample amount.
This time around, we're actually getting to use the entirety of this formula. I'll zip past the first part:
3.4x10^{24 }/ 6.02x10^{23} = 5.65 moles
Okay, this is where we pull the second half of the equation:
Mol x m_{base formula} = m_{sample amount }
We have the moles, 5.65, what we need now is the base formula mass to get the mass of the sample amount.
N = 14.007
H_{3} = 3.024
add it together to get 17.031g.
5.65 x 17.031 = m_{sample amount }
= 96.22515g Easy
5) How much does 4.2 moles of Ca(NO_{3})_{2} weigh?
Another one like this? Well, luckily the "weigh" part hints that we're finding the sample mass yet again. We can axe out the 1st part of The Equation, and now we can get onto to the juicy part.
Mol x m_{base formula} = m_{sample amount }Remember, we have the moles and we need the base formula.
Ca = 40.078
N_{2} = 14.007 x 2 = 28.014
O_{6} = 15.999 x 6 = 95.994
which equals 164.312 (I'm not making this up).
4.2 x 164.312 = 690.1104g
6) What is the molar mass of MgO?
Easy question much? Just look on your periodic table, locate Mg and O's atomic mass (which, by the way, are 24.305 and 15.999, respectively) and add 'em together. Bam, molar mass equals 40.304g.
7)How are the terms "molar mass" and "atomic mass" different from one another?
Atomic mass is the mass of an atom. Molar mass is the mass of the moles of a substance.
8) Which is a better unit for expressing molar mass, "amu" or "Grams/mole"?
Grams/mole is better, I am in thinking. ;)
1) Define "mole".
Mole is a unit of measure for a substance.
2) How many moles are present in 34 grams of Cu(OH)_{2}?
To find out how many moles are present in X amount of anything, remember that Moles x Formula Mass equals the grams. So, find out the mass of Cu(OH)_{2} which isCu: 63.543
O_{2}: 15.999*2 : 31.998
H_{2}: 2.106
Which equals 97.647. So we have the formula mass (I'll call it F) and we have the given mass (34, which I'll call G). Moles I'll call M.
M x F = G
97.647M = 34
Working backwards, we say 34/97.647 = about .35 moles.
3) How many moles are present in 2.45x10^{23} molecules of CH_{4}?
Recall fromMole Potpourri and other sources that
P/A = Mol
Mol x m_{base formula} = m_{sample amount}
Or in other words, The particle amount divided by A's # = the moles AND
Moles times mass of base formula = mass of the sample amount.
Luckily, we're only trying to get to the moles, so discard the second equation [for now].
We know the particle amount of CH_{4} sample is 2.45x10^{23}. And we also know A's number is 6.02x10^{23}. Divide 2.45x10^{23} by 6.02x10^{23} and you'll get approximately .41 moles.
4) How many grams are there in 3.4x10^{24} molecules of NH_{3}?
Recall fromMole Potpourri and other sources that
P/A = Mol
Mol x m_{base formula} = m_{sample amount}
Or in other words, The particle amount divided by A's # = the moles AND
Moles times mass of base formula = mass of the sample amount.
This time around, we're actually getting to use the entirety of this formula. I'll zip past the first part:
3.4x10^{24 }/ 6.02x10^{23} = 5.65 moles
Okay, this is where we pull the second half of the equation:
Mol x m_{base formula} = m_{sample amount }
We have the moles, 5.65, what we need now is the base formula mass to get the mass of the sample amount.
N = 14.007
H_{3} = 3.024
add it together to get 17.031g.
5.65 x 17.031 = m_{sample amount }
= 96.22515g Easy
5) How much does 4.2 moles of Ca(NO_{3})_{2} weigh?
Another one like this? Well, luckily the "weigh" part hints that we're finding the sample mass yet again. We can axe out the 1st part of The Equation, and now we can get onto to the juicy part.
Mol x m_{base formula} = m_{sample amount }Remember, we have the moles and we need the base formula.
Ca = 40.078
N_{2} = 14.007 x 2 = 28.014
O_{6} = 15.999 x 6 = 95.994
which equals 164.312 (I'm not making this up).
4.2 x 164.312 = 690.1104g
6) What is the molar mass of MgO?
Easy question much? Just look on your periodic table, locate Mg and O's atomic mass (which, by the way, are 24.305 and 15.999, respectively) and add 'em together. Bam, molar mass equals 40.304g.
7)How are the terms "molar mass" and "atomic mass" different from one another?
Atomic mass is the mass of an atom. Molar mass is the mass of the moles of a substance.
8) Which is a better unit for expressing molar mass, "amu" or "Grams/mole"?
Grams/mole is better, I am in thinking. ;)
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