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Chapter 3: Problem 35

A compound of carbon and hydrogen consists of \(94.34 \% \mathrm{C}\) and \(5.66 \% \mathrm{H},\) by mass. The molecular mass of the compound is found to be 178 u. What is its molecular formula?

Short Answer

Expert verified
The molecular formula for the given carbon and hydrogen compound is \(C12H18\).

Step by step solution

01

Convert the percentage composition into grams

Since percentage is based out of 100, for simplicity we can consider we have 100 g of the compound. Therefore, if \(94.34 \% \mathrm{C}\) by mass, assume 94.34 g of Carbon. If \(5.66 \% \mathrm{H}\) then assume 5.66 g of Hydrogen.
02

Calculate the ratio of moles of Carbon and Hydrogen to find the empirical formula

Now, calculate the number of moles for each element. The number of moles = Mass of the element / molar mass of the element. Carbon: 94.34 g / 12.01 g/mol = 7.85 mol Carbon. Hydrogen: 5.66 g / 1.008 g/mol = 5.61 mol Hydrogen. After this, divide each of the calculated moles by the smallest quantity of moles calculated, to get the molar ratio. Carbon: 7.85 mol / 5.61 = 1.40 mol. Hydrogen: 5.61 mol / 5.61 = 1 mol. So, the empirical formula is \(CH_{1.4}\). But we only use integers in subscripts, and 1.4 is very close to 1.5, which when multiplied by 2 gives a whole number. Therefore, Multiply all the subscripts by 2 to convert them to whole numbers. This gives the Empirical formula as \(C2H3\).
03

Determine the molecular formula

The molecular formula is a multiple of the empirical formula. The molar mass of the empirical formula \(C2H3\) is 29 g/mol. Given the molecular mass is 178 u. Therefore, multiple = Molecular mass / Empirical formula mass = 178/29 = 6.13 ≈ 6. So, the molecular formula = 6 * Empirical formula = \(C12H18\).

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Most popular questions from this chapter

Three of the following formulas might be either an empirical or a molecular formula. The formula that must be a molecular formula is (a) \(\mathrm{N}_{2} \mathrm{O} ;\) (b) \(\mathrm{N}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{NaCl} ;\) (d) \(\mathrm{NH}_{3}\)

A 0.622 g sample of a metal oxide with the formula \(\mathrm{M}_{2} \mathrm{O}_{3}\) is converted to \(0.685 \mathrm{g}\) of the sulfide, MS. What is the atomic mass of the metal M?

The food flavor enhancer monosodium glutamate (MSG) has the composition \(13.6 \%\) Na, \(35.5 \%\) C. \(4.8 \% \mathrm{H}, 8.3 \% \mathrm{N}, 37.8 \% \mathrm{O},\) by mass. What is the empirical formula of MSG?

All of the following minerals are semiprecious or precious stones. Determine the mass percent of the indicated element. (a) \(\mathrm{Zr}\) in zircon, \(\mathrm{ZrSiO}_{4}\) (b) Be in beryl (emerald), \(\mathrm{Be}_{3} \mathrm{Al}_{2} \mathrm{Si}_{6} \mathrm{O}_{18}\) (c) Fe in almandine (garnet), \(\mathrm{Fe}_{3} \mathrm{Al}_{2} \mathrm{Si}_{3} \mathrm{O}_{12}\) (d) S in lazurite (lapis lazuli), \(\mathrm{Na}_{4} \mathrm{SSi}_{3} \mathrm{Al}_{3} \mathrm{O}_{12}\)

The oxidation state of I in the ion \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\) is (a) -1 (b) \(+1 ;(c)+7 ;(d)+8\)
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