<p>1)A sample of polystyrene prepared by heating styrene with tribromobenzoyl peroxide in the absnece of air as the formula Br3C6H3(C8H8)n. The number n varies with the conditions of the preparation. One sample of polystyrene prepared in this manner was found to contain 10.46% bromine. What is the value of n?</p>
<p>2) CaC2 is made in an electric furnace by the reaction CaO + 3C > CaC2 +CO. How much CaO is to be added to the furnace charge for each 40gram of CaC2 produced?</p>
<p>3)Calculate the mass of KCLO3 necessary to produce 1.23g of O2. What mass of KCL is produced along with this quantity of oxygen?</p>
<p>4)Elementary analysis showed that an organic compound contained C,H,N, and O as its only elementary constituents. A 1.279-g sample was burned completey, as a result of whic 1.60g of CO2 and .77g of H20 were obtained. A seperately weighed 1.625 g sample contained .216g nitrogen. What is the empirical formula?</p>
<p>for the last one, I know what to do with the C, H, and O....but what do I do with the N. The formula I got was C7H17O9N3......but i'm pretty sure this is wrong. </p>
<p>PLEASE HELP!!!</p>
<p>1.</p>
<p>The amount of bromine is 10.46% of the total weight of the sample. Now put this into mathematical terms.</p>
<p>239.72 g Br = 1046/10000 <em>(239.72 g of Br + 72.06 g C + 3.024 g H + n</em>[16.08 g C + 8.064 g H] )</p>
<p>Now, solve for n. You should get something like 81.88. Then you can ound it to 82, since n should be an integer.</p>
<p>2.</p>
<p>Use stoichiometry.</p>
<p>40 g CaC2 x (1 mol CaC2 / # g CaC2) x (1 mol CaO/ 1 mol CaC2) x (# g CaO / 1 mol CaO) = ANSWER.</p>
<p>Use percentage for Nitrogen. You know that 0.216 g / 1.625 g = % composition of Nitrogen.</p>
<p>1.assume 100 g, so 10.46 g Br (1 mol Br/79.9g Br) =0.13091 mols Br
0.13091 mols Br (3 mols C6H3(C8H8)n)/1 mol Br)= 0.392739 mols C6H8H3(C8H8)n
100-10.46=89.54 g C6H3(C8H8)n ..g/mol =89.54/0.392739=227.98 g/mol
227.98-(6X12 +1X3)=152.98
12X8 + 1X8= 104
Yeah, did something wrong there, but I think that is the basic idea..<em>confused</em></p>
<ol>
<li><p>40g CaC2 (1mol CaC2/111 g CaC2)(1 mol Ca0/1mol CaC2)(56g Ca0/1 mol Ca0)=20.18 g Ca0 needed.</p></li>
<li><p>KCL03 -->O2 + KCL..balance...2KCLO3--->3O2 +2KCl
1.23 g O2 (1 mol O2/32g 02)(2 mol KCLO3/3 mol 02)(122.55g KClO3/1 mol KCLO3)=3.14 g KCLO3</p></li>
</ol>
<p>3.14 g KCLO3(1 mol KCLO3/122.55 g KCLO3)(2 mol KCL/2 mol KCLO3)(74.55g KCl/1 mol KCL)=1.91 g KCl</p>
<p>.216 g/1.625 g sample N=13.292 % N
12X7+17+16X9=245 g
N=14 g/mol
and 3X14=42 42/245+42)=14.6%..um, thats sort of close, so I would say that 3 is right..</p>
<p>Sorry I couldn't help more!</p>
<p>Hah, took so long to write that out, people already gave you answers</p>
<p>^ you guys are the greatest.....thanx</p>
<p>@ lil_killer129......regarding post #3: what do I do with the %comp of N....convert it into moles? for instance, say i get 20%, do I then say 20g /14g/mol ?</p>
<p>Yes. That's what you do. After you have the % composition for each element, you can then assume 100 grams and proceed from there.</p>
<p>oh wait.....i still need some clarification on #'s 1 and 4, please.</p>
<p>For number 1, it says that Br is 10.46% of the total weight. You know the total weight of Br in the sample (which is 3x weight of 1 Br or 239.72 g Br). The word "is" in mathematical terms means "equal" or the symbol "=" if you prefer. 10.46% is 10.46/100 or 1046/10000 (multiplied by 100/100 to clear the decimals). The word "of" in mathematical terms means to multiply or the symbol "x" if you prefer. The total weight is just the sum of the weight of each individual element in the sample; so you would simply "sum up" the weights. Your sample has 3 Br's (3 x mass of Br), 6 C's (6 x mass of C), 3 H's (3 x mass of H), n C8H8's (n x mass of C8H8).</p>
<p>Thats how you get this formula:</p>
<p>(Mass of Br in sample = % <em>TOTAL MASS)
239.72 g Br = 1046/10000 *(239.72 g of Br + 72.06 g C + 3.024 g H + n</em>[16.08 g C + 8.064 g H] )</p>
<p>Now solve for n (which is the number of C8H8's in the sample). You can do a quick check by taking the weight of Br (239.72 g) in your entire sample and divide that by your total weight to see if you get 10.46% Br.</p>
<p>I hope this helps.</p>