Assuming that it is sucrose(table sugar), it has the formula C12H22O11, or a molar mass of 342 g/mol. 6/342=0.0175 moles.
2007-07-02 11:18:48
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answer #1
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answered by sword_kid 2
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You should have stated the question "How many moles of sugar are in 1 gram of sugar" then multiplied that by 6. I am thinking the number would be a lot bigger than you would expect though. I'll look into it too and if I find anything I'll update my answer.
2007-07-02 11:21:04
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answer #2
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answered by mute8s 2
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You need to find out what kind of sugar (fructose, glucose, sucrose), There might be more than one. If there is more than one you'd have to know the percentages of each but I doubt its that complicated. Find the molar mass of the sugar in question then do the math to find moles.
2007-07-02 11:13:29
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answer #3
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answered by Lady Geologist 7
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moles = grams of substance * 1 mol/grams of substance from molar mass
to convert from grams to moles, use molar mass, to convert from moles to grams, use molar mass
the molar mass = sum of all of the masses of every atom in the substance
For example, the molar mass of water = H2O
H: 2 x 1.01 (mass off periodic table) = 2.02 grams/mol
O: 1 x 16.00 = 16.00 grams/mol
molar mass = 16.00 + 2.02 = 18.02 grams/mol
Let's say you have 6 grams of water
6 grams H2O x 1 mole H2O/18.02 grams = 0.332 moles
You need to know what sugar you are talking about so you can calculate the molar mass of that sugar (sucrose? fructose? glucose?) and then use the 6 grams sugar x 1 mol sugar/molar mass sugar
2007-07-02 12:23:23
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answer #4
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answered by Anonymous
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I would think grams to moles... but you need to know the molar mass of sugar... eh... I don't like chemistry... or at least not the stoichiometry(sp?) aspect of it.
2007-07-02 11:11:23
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answer #5
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answered by Anonymous
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Ahh Chemistry, I still don't have a clue what it is about but, have a look and see if it helpls
2007-07-02 11:16:25
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answer #6
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answered by Anonymous
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Mick Dale and Dave Lewis posted the same question. You should see the answers side by side.
2016-08-24 07:29:26
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answer #7
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answered by renae 4
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