Heavy hydrocarbons

6561 /www/nrich/html/content/id/6561/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <ul id="stemLinks"> <li><a href="http://nrich.maths.org/920">Warm-up</a></li> <li><a href="http://nrich.maths.org/6562">Try this next</a></li> <li><a href="http://nrich.maths.org/6581">Think higher</a></li> <li><a href="http://nrich.maths.org/5904">Read: mathematics</a></li> <li><a href="http://en.wikipedia.org/wiki/Hydrocarbon">Read: science</a></li> <li><a href="http://en.wikipedia.org/wiki/Isotope">Explore further</a></li> </ul> <div> </div> This problem consists of a series of questions concerning hydrocarbons made from the stable isotopes of hydrogen and carbon. Assume that the probability of the occurrence of unstable isotopes is effectively zero. Assume also that the mass of an isotope is equal to the mass number. Carbon is found in two naturally occurring stable isotopic forms: Carbon-12, $^{12}$ C, accounts for $98.9\%$ of naturally occurring carbon, and Carbon-13, $^{13}$ C, accounts for $1.1\%$ of naturally occurring carbon. Hydrogen also has two stable isotopes: $^1$H accounts for slightly more than $99.985\%$ of naturally occurring hydrogen. $^2$H, called deuterium, accounts for about $0.015\%$ of naturally occurring hydrogen. What possibilities are there for the molecular mass of a stable molecule of methane CH$_4$? Which three are the most common? What is the probability that a randomly encountered methane molecule will be of each of these masses? What would be the three most likely possibilities for the molecular masses of the next two alkanes: ethane, C$_2$H$_6$, and propane C$_3$H$_8$? Am I likely to find a molecule of butane C$_4$H$_{10}$ of molecular mass $72$ in a litre of butane? (Avogadro's constant is approximately $6.02\times 10^{23}$). Are there any alkanes for which the most commonly found molecule is not the one made entirely from $^{12}$C and $^1$H? <div class="framework"> NOTES AND BACKGROUND Mass spectrometry allows chemists to determine the molecular masses of compounds. The frequency of occurrence of common isotopes of elements is an important factor in understanding such as analysis. Mass spectrometry is of great industrial and scientific importance, and in practice, the results of an analysis are determined by comparison with a database of experimental spectrometry information. For information on alkanes and hydrocarbons, see <a href="http://en.wikipedia.org/wiki/Hydrocarbon">http://en.wikipedia.org/wiki/Hydrocarbon</a></div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Note that in actuality the masses of the different isotopologues of $\text{CH}_4$ are slightly different. These differences may be noted by a very sensitive mass spectrometer. Take for example: RMM $^{12}\text{CH}_3\text{D}$ = 12 + 3(1.007825) + 2.014102 = 17.037577 gmol$^{-1}$ RMM $^{13}\text{CH}_4$ = 13.00335 + 4(1.007825) = 17.03465 gmol$^{-1}$ where the calculation is limited by the degree of accuracy of the given data. However in this question it is acceptable to take values to the nearest gmol$^{-1}$, giving roughly equal molecular masses for certain isotopes. $^{12}\text{CH}_4$ = 16 gmol$^{-1}$ $^{12}\text{CH}_3\text{D}$, $^{13}\text{CH}_4$ = 17 gmol$^{-1}$ $^{12}\text{CH}_2\text{D}_2$, $^{13}\text{CH}_3\text{D}$ = 18 gmol$^{-1}$ $^{12}\text{CHD}_3$, $^{13}\text{CH}_2\text{D}_2$ = 19 gmol$^{-1}$ $^{12}\text{CD}_4$, $^{13}\text{CHD}_3$ = 20 gmol$^{-1}$ $^{13}\text{CD}_4$ = 21 gmol$^{-1}$ The three most common species encountered would be $^{12}\text{CH}_4$, $^{13}\text{CH}_4$ and $^{12}\text{CH}_3\text{D}$ in order of likelihood. This can bee seen intuitively as the probability of encountering a $^{13}\text{C}$ more likely than encountering a single $^{2}\text{H}$, and for a small molecule such as methane, it is far more likely to obtain $^{12}\text{CH}_4$ than either of the other possibilities. The three most likely molecular masses as 16, 17 and 18 gmol$^{-1}$ as the introduction of more $^{13}\text{C}$ and $^{2}\text{H}$ to a small molecule reduces its likelihood more than any combinatorial effects can compensate for. The actual probabilities of encountering each of these molecular masses of methane are: 16 gmol$^{-1}$ [$^{12}\text{CH}_4$] $\textbf P(16 \text{gmol}^{-1}) = 0.989 \times (0.99985)^4 = 0.988$ (3sf) 17 gmol$^{-1}$ [$^{12}\text{CH}_3\text{D}$, $^{13}\text{CH}_4$] $\textbf P(17 \text{gmol}^{-1}) = (0.989\times^4\textbf{C}_1(0.00015)\times (0.99985)^3 ) + (0.011 \times (0.99985)^4) = 0.0116$ (3sf) 18 gmol$^{-1}$ [$^{12}\text{CH}_2\text{D}_2$, $^{13}\text{CH}_3\text{D}$] $\textbf P(18 \text{gmol}^{-1}) = (0.989\times^4\textbf{C}_2(0.00015)^2\times (0.99985)^2 ) + (0.011\times^4\textbf{C}_1(0.00015)\times (0.99985)^3) = 6.73\times10^{-6}$ (3sf) 19 gmol$^{-1}$ [$^{12}\text{CHD}_3$, $^{13}\text{CH}_2\text{D}_2$] $\textbf P(19 \text{gmol}^{-1}) = (0.989\times^4\textbf{C}_3(0.00015)^3\times 0.99985 ) + (0.011\times^4\textbf{C}_2(0.00015)^2\times (0.99985)^2) = 1.50\times10^{-9}$ (3sf) 20 gmol$^{-1}$ [$^{12}\text{CD}_4$, $^{13}\text{CHD}_3$] $\textbf P(20 \text{gmol}^{-1}) = (0.989\times^4\textbf{C}_4(0.00015)^4) + (0.011\times^4\textbf{C}_3(0.00015)^3\times 0.99985) = 1.49\times10^{-13}$ (3sf) 21 gmol$^{-1}$ [$^{13}\text{CD}_4$] $\textbf P(21 \text{gmol}^{-1}) = (0.011\times^4\textbf{C}_4(0.00015)^4) = 5.57\times10^{-18}$ (3sf) Following a similar principle to that above, the three most likely possibilities for the molecular masses of ethane are 30, 31 and 32 gmol$^{-1}$ in order of likelihood. $^{12}\text{C}_2\text{H}_6$ = 30 gmol$^{-1}$ $^{12}\text{CD}_2\text{H}_5$, $^{12}\text{C}^{13}\text{C}\text{H}_6$ = 31 gmol$^{-1}$ $^{12}\text{C}_2\text{H}_4\text{D}_2$, $^{12}\text{C}^{13}\text{CD}\text{H}_5$, $^{13}\text{C}_2\text{H}_6$ = 32 gmol$^{-1}$ $\textbf P(30 \text{gmol}^{-1}) = (0.989)^2 \times (0.99985)^6 = 0.977$ (3sf) $\textbf P(31 \text{gmol}^{-1}) = ((0.989)^2 \times ^6\textbf{C}_10.00015 \times (0.99985)^ 5) + ((0.989)\times^2\textbf{C}_1(0.011)\times(0.99985)^ 6) = 0.0226$ (3sf) $\textbf P(32 \text{gmol}^{-1}) = ((0.989)^2 \times ^6\textbf{C}_2(0.00015)^2 \times (0.99985)^ 4) + ((0.989)\times^2\textbf{C}_1(0.011)\times^6\textbf{C}_1(0.00015)\times(0.99985)^ 5)$ $+ (^2\textbf{C}_2(0.011)^2 \times (0.99985)^6) = 1.41 \times 10^{-4} \text{ (3sf)}$ The three most likely possibilities for the molecular mass of propane are 44, 45 and 46 gmol$^{-1}$ in order of likelihood. $^{12}\text{C}_3\text{H}_8$ = 44 gmol$^{-1}$ $^{12}\text{C}_2{}^{13}\text{C}\text{H}_8$, $^{12}\text{C}_3\text{D}\text{H}_7 = 45\text {gmol}^{-1}$ $^{12}\text{C}^{13}\text{C}_2\text{H}_8$, $^{12}\text{C}_2^{13}\text{C}\text{D}\text{H}_7, ^{12}\text{C}_3\text{D}_2\text{H}_ 6 = 46\text {gmol}^{-1}$ $\textbf P(44 \text{gmol}^{-1}) = (0.989)^3 \times (0.99985)^8 = 0.966$ (3sf) $\textbf P(45 \text{gmol}^{-1}) = (^3\textbf{C}_1 \times 0.011 \times (0.989)^2 \times (0.99985)^ 8) + ((0.989)^3\times^8\textbf{C}_1(0.00015)\times(0.99985)^ 7) = 0.0334$ (3sf) $\textbf P(46 \text{gmol}^{-1}) = (^3\textbf{C}_2 \times (0.011)^2 \times 0.989 \times (0.99985)^ 8) + (^3\textbf{C}_1 \times 0.011 \times (0.989)^2 \times^8\textbf{C}_1(0.00015)\times(0.99985)^ 7)$ $+ ((0.989)^3\times^8\textbf{C}_2(0.00015)^2\times(0.99985)^ 6) = 3.98 \times 10^{-4}$ (3sf) A molecule of butane with molecular mass 72 is the isotopologue $^{13}\text{C}_4\text{D}_{10}$. The probability of any butane molecule being this isotopologue is: $\textbf{P}(^{13}\text{C}_4\text{D}_{10}) = 0.011^4 \times 0.00015^{10} = 8.44 \times 10^{-47}$ (3sf) 24dm$^3$ of butane corresponds to roughly 1 mole of butane molecules. Thus, as 1 dm$^3$ is equivalent to a litre, the number of moles in the required volume is $\frac{1}{24}$. The number of molecules in the sample is given by multiplying the number of moles, by the number of molecules in a mole (the Avogadro's constant): Number of molecules $= \frac{N_A}{24} = 2.51 \times 10^{22}$ So the likelihood can be found by multiplying the probability for one molecule by the total number of molecules in the sample. $\therefore\text{E}(^{13}\text{C}_4\text{D}_{10}) = 2.51 \times 10^{22} \times 8.44 \times 10^{-47} = 2.12 \times 10^{-24}$ This question requires a little more algebraic appreciation of the calculations thusfar. The probability of encountering a generic alkane $^{12}\text{C}_n{}^1\text{H}_{2n + 2}$ is given by: $\textbf{P} = 0.989^n \times (0.99985)^{2n + 2}$ The probability of encountering an isotopologue containing deuterium $^{12}\text{C}_n^{\ 1}\text{H}_{2n + 1}\text{D}$ is given by: $\textbf{P} = 0.989^n \times ^{2n + 2}\textbf{C}_1(0.99985)^{2n + 1}\times 0.00015$ The probability of encountering this isotopologue must be greater than the likelihood of finding the butane molecule consisting entirely of $^{12}\text{C}$ and H. $0.989^n \times ^{2n + 2}\textbf{C}_1 \times (0.99985)^{2n + 1}\times 0.00015$ > $0.989^n \times (0.99985)^{2n + 2}$ $0.00015 (2n + 2)$ > 0.99985 2n + 2 > 6665.666667 n > 3331.833333 n = 3332 Is this long chain alkane likely to exist in a real sample? The probability of encountering an isotopologue containing $^{13}\text{C}$ is given by: $\textbf{P} = ^n\textbf{C}_1\times 0.011 \times 0.989^{n-1} \times (0.99985)^{2n + 2}$ This probability must be greater than the likelihood of finding the alkane molecule consisting entirely of $^{12}\text{C}$ and H. So: $^n\textbf{C}_1\times 0.011 \times (0.989)^{n-1} \times (0.99985)^{2n + 2}$ > $0.989^n \times (0.99985)^{2n + 2}$ $ \text{n} \times (0.989)^{n-1} \times 0.011$ > $(0.989)^{n}$ $0.011\text{n}$ > 0.989 $\text{n}$ > 89.90909091 n = 90 Consider the likelihood of the existence of this molecule. As an extension, does the probability of the existence of such molecules change if the molecule is produced via a method which involves polymerisation? Try to construct an algebraic test. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><a href="http://nrich.maths.org/6561">This activity</a> makes a good homework or individual task. It uses GCSE probability and A-level chemistry ideas, and gives a good exercise in combinatorics in an interesting context. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Remember that molecules with more than one hydrogen and carbon atom can have isotopes at different places. For example, in ethane, a carbon-13 could occur at either of the carbon sites! </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Carbon the ratios are almost exactly 90:1 Hydrogen, the ratios are almost 5500:1 </mdoxml> 2 3 0 0 0 1 0 Heavy hydrocarbons Explore the distribution of molecular masses for various hydrocarbons Probability Combining probabilities Applications chemistry Probability Probability Applications chemistry Admin Individual Applications STEM - physical world