Chemistry Help

1. Sep 5, 2007

Destrio

Problem 1:

The class of ternary compounds called carboranes (containing the elements carbon, boron and hydrogen) have played a prominent role in many areas of fundamental chemical research. In particular, they have provided many interesting examples for testing theories on chemical bonding and structure.
A particular carborane has the following mass percentages: 59.28%B and 14.37%H.
What is the empirical formula of this compound?

I tried reducing them, getting 19.76 , 4.79, and 18.68 , but it was no help to me as I have never worked with mass percentages before and am lost on where to proceed. (My only hint was to find the GCF)

Problem 2:

suppose a nameless element (X) was found to occur in three different substances (A, B and C). The relative molecular masses of A, B, and C were found to be 27.40, 52.90 and 62.60 respectively, compared to an assigned value of exactly 2.00 for H2. Subsequently, these substances were subjected to elemental analysis and were found to contain 78.1%, 81.1% and 85.6% of X respectively by mass.

What is the (likely) relative atomic mass of X, on this scale?

Same as above, I found the masses to be 21.4 , 42.9 , and 53.6 individually

Problem 3:

A white powder, consisting of a simple mixture of tartaric acid (C4H6O6) and citric acid (C6H8O7) was analysed to determine the elemental composition. Combustion of a 387.2-mg sample produced 502.2 mg of CO2 and 143.0 mg of H2O.

Use atomic masses: C 12.011; H 1.00794; O 15.9994.

Calculate the % carbon, by mass, in the sample.

I found divided the molar mass of carbon in mg but the amount of CO2 produced and got .04181% which didn't seem right, and isn't correct. I'm not sure how to begin this question.

2. Sep 5, 2007

atavistic

For part a) Take the 100g sample and then get the no. of moles of each and then find the simplest ratio.

For part b ) The equation should be like this : (78.1*27.4 + 52.9*81.1 + 62.6*85.6)/100 as this is how you find the relative weight.

PArt c) I need to thnk a little more.

3. Sep 8, 2007

rocomath

q3 - write 2 separate combustion rxns for tartaric acid and citric acid, then combine the 2 rxn eqns and go from there

edit: after trying to balance the rxns separately, i found that just writing it all in 1 is easier.

i'm working on it right now, i've had a similar problem and it's the same way i did it; let me check tho.

i'm lost on this problem ... i need to think a little more too, lol.

Last edited: Sep 8, 2007
4. Sep 8, 2007

chemisttree

You have a compound that is composed of C, H and B. You are given percentages by weight for two of these elements and you must determine the third. Once you have the percentage by weight of each of the three elements, divide them by their atomic mass (use the mass accurate to +/-0.001 g/mol). Divide them all by the smallest mole # you get and then multiply each by an integer to get three approximately integer values.

For example if you get for C, H, B the molar values 3.324, 4.992, and 2.497, you would first divide all by 2.497. The results would be 1.331, 1.999 and 1.000. Multiplying these by '3' yields 3.993, 5.998 and 3.000 which would give an empirical formula of $$C_4H_6B_3$$. Notice I use the +/- 0.001 precision throughout to minimize errors. I did that because the precision is one greater than the precision given in the original problem (59.28% B and 14.37 H).

The second problem is done the same way.

The third problem is a little tricky because too much information is given to you. All you need to know is that each carbon will produce a CO2 molecule and that the mass of the sample is 387.2 mg and that 502.2 mg of CO2 is produced. All the other information is noise.

5. Sep 8, 2007

rocomath

i figured i was doing too much work when i had my rxn written out. the problem already states told how much sample there is and how much CO2 is produced.

6. Sep 11, 2007

Destrio

thanks, i was able to get the first one by dividing the percentages by the atomic masses
but for the 2nd i divided the %s by the given molecular masses, 78.1/27.40 ... etc
getting 2.85, 1.53 and 1.37
i divided all those by 1.37 getting 1, 1.12, and 2.08
1 as relative atomic masses didn't work, neither did multiplying them all by 10
what have i done wrong here?

thanks