# Percentage Yield

For the above MCQ question, I took the equation to be $$\ce{CaC2 + 2H2O -> C2H2 + Ca(OH)2}$$ I then calculated the moles of $$\ce{CaC2}$$ as 0.7273 moles, which has a 1:1 molar ratio to $$\ce{C2H2}$$, making the theoretical mass of $$\ce{C2H2}$$ 0.7273 x 26 = 18.9 g

The % yield = (11.5/18.9) x 100 = 60.8%.

However, it is not an MCQ option. What am I doing wrong?

• Please cite the source of the quoted problem. People deserve to know what trash textbooks with zero copy-editing they should avoid. Also, the text should be text, not an image. The title should be adequate to distinguish the question without opening the main body. Commented May 17 at 12:48
• Sorry about that. This is from a Junior Science Olympiad paper. Commented May 19 at 7:30
• Note kg, not g, but that does not affect the ratio. 32 kg CaC2 =730 moles; 11.5 kg C2H2 = 440 moles, 440/730 = ~61%. Your reaction is correct, your math checks, and yet 61% is not a choice... ask the instructor, as it might be an error in the text. (Not sure why this question was down-voted; work is shown). Commented May 31 at 3:41

Let:

A = $$\ce{CaC2}$$

B = $$\ce{C2H2}$$

$$\mathrm{yield}=\frac{m_\mathrm{B}}{m_\mathrm{B}^*}=\frac{m_\mathrm{B}}{n_\mathrm{B}^*\;M_\mathrm{B}}=\frac{m_\mathrm{B}}{n_\mathrm{A}\;M_\mathrm{B}}=\frac{m_\mathrm{B}}{\frac{m_\mathrm{A}}{M_\mathrm{A}}\;M_\mathrm{B}}=\frac{\pu{11.5kg}}{\left(\frac{\pu{32kg}}{\pu{64kg/kmol}}\right)\pu{26kg/kmol}}=0.88$$

Note: The molar mass of calcium is approximately $$\pu{40kg/kmol}$$, not $$\pu{20kg/kmol}$$ as the problem statement says.

Note: Variables with an asterisk are in reference to the theoretical values.

• What does asterisk refer to? Personally I wouldn't go with substitution: the formula is already quite simple and C = $\ce{C2H2}$ looks pretty confusing in a chemical setting. Do not arbitrarily switch between upright and italic notations, this is also perplexing. Strictly speaking, $0.88 \neq 88\%.$ Also, MathJax is not suited for emphasis. Commented May 17 at 12:45
• Updated to reflect given observations. Although I'm not understanding the claim: $0.88\neq 88\%$ Commented May 17 at 17:44
• Percent is a unit, whereas the fraction for the yield is unitless. An equal sign would be only justified for $0.88\times100\% = 88\%.$ Since you probably don't want to multiply every step by $100\%,$ I'd suggest to use $\ldots = 0.88~\mathrm{or}~88\%.$ And you still got italicized and upright As and Bs mixed up in the substitution and equation steps. Commented May 17 at 18:20
• Variables are italicized, textual labels (which, BTW, 'yield' is as well) are upright. For further guidance and reference, see for instance section 1.6 in IUPAC Green Book. For chemical expressions use \ce{…}, for upright text in math mode in general — \mathrm{…}. Same as with $\mathrm\LaTeX$ and across STEM in general. Commented May 17 at 19:17
• I see. I Will keep in mind these details in future answers. Thank you. Commented May 17 at 19:33