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Chapter 3: Problem 45

Dimethylhydrazine is a carbon-hydrogen-nitrogen compound used in rocket fuels. When burned in an excess of oxygen, a \(0.312 \mathrm{g}\) sample yields \(0.458 \mathrm{g} \mathrm{CO}_{2}\) and \(0.374 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). The nitrogen content of a \(0.486 \mathrm{g}\) sample is converted to \(0.226 \mathrm{g} \mathrm{N}_{2} .\) What is the empirical formula of dimethylhydrazine?

Short Answer

Expert verified
The empirical formula of Dimethylhydrazine is \( \mathrm{CH}_{4}\mathrm{N}_{2} \).

Step by step solution

01

Find moles of Carbon

The carbon that formed in the compound is now in the form of \( \mathrm{CO}_{2} \). One mole of \( \mathrm{CO}_{2} \) contains one mole of carbon. So, using the given mass of the \( \mathrm{CO}_{2} \), the moles of carbon can be found by dividing the mass by the molar mass of \( \mathrm{CO}_{2} \). In this case, that would be \(0.458 / 44.01 = 0.0104 \) moles.
02

Find moles of Hydrogen

The hydrogen that was present in the compound is now in the form of \( \mathrm{H}_{2}\mathrm{O} \). Each mole of \( \mathrm{H}_{2}\mathrm{O} \) contains two moles of hydrogen. So, the moles of hydrogen can be found by first finding the moles of \( \mathrm{H}_{2}\mathrm{O} \) (0.374 / 18.015) and then multiplying by 2 to get the number of moles of hydrogen. This equation comes to about \(0.0414 \) moles.
03

Find moles of Nitrogen

The nitrogen content is present as \( \mathrm{N}_{2} \), one mole of which contains two moles of nitrogen. So, in order to find the moles of nitrogen, the weight of the \( \mathrm{N}_{2} \) is divided by the molar mass of nitrogen and then multiplied by 2. Here, that is \( (0.226 / 28.02) * 2 = 0.0161 \) moles.
04

Determine the Empirical Formula

Normalize the obtained moles to get simplest whole number ratio, by dividing each value by the smallest value among the three. Here, that would mean dividing \(0.0104, 0.0414 \), and \(0.0161 \) by \(0.0104 \). This results in a mole ratio of \( C:1, H:4, N:2 \) denoting the empirical formula to be \( \mathrm{CH}_{4}\mathrm{N}_{2} \)
05

Confirm the Answer

The empirical formula determines the simplest whole-number ratio of atoms in a compound. In this case, \( \mathrm{CH}_{4}\mathrm{N}_{2} \) reflects a compound that is composed of 1 atom of carbon, 4 atoms of hydrogen, and 2 atoms of nitrogen. Therefore, the empirical formula of Dimethylhydrazine is \( \mathrm{CH}_{4}\mathrm{N}_{2} \).

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Most popular questions from this chapter

All-purpose fertilizers contain the essential elements nitrogen, phosphorus, and potassium. A typical fertilizer carries numbers on its label, such as "5-10-5". These numbers represent the \% \(\mathrm{N}, \% \mathrm{P}_{2} \mathrm{O}_{5},\) and \(\% \mathrm{K}_{2} \mathrm{O},\) respectively. The \(\mathrm{N}\) is contained in the form of a nitrogen compound, such as \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}, \mathrm{NH}_{4} \mathrm{NO}_{3}\) or \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\) (urea). The \(\mathrm{P}\) is generally present as a phosphate, and the \(K\) as \(K C\). The expressions \(\% \mathrm{P}_{2} \mathrm{O}_{5}\) and \(\% \mathrm{K}_{2} \mathrm{O}\) were devised in the nineteenth century, before the nature of chemical compounds was fully understood. To convert from \% \(\mathrm{P}_{2} \mathrm{O}_{5}\) to \% \(\mathrm{P}\) and from \% \(\mathrm{K}_{2} \mathrm{O}\) to \% \(\mathrm{K}\), the factors \(2 \mathrm{mol} \mathrm{P} / \mathrm{mol}\) \(\mathrm{P}_{2} \mathrm{O}_{5}\) and \(2 \mathrm{mol} \mathrm{K} / \mathrm{mol} \mathrm{K}_{2} \mathrm{O}\) must be used, together with molar masses. (a) Assuming three-significant-figure precision, what is the percent composition of the "5-10-5" fertilizer in \% \(\mathrm{N}, \% \mathrm{P},\) and \(\% \mathrm{K} ?\) (b) What is the \(\% \mathrm{P}_{2} \mathrm{O}_{5}\) in the following compounds (both common fertilizers)? (i) \(\mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}\) (ii) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\) (c) In a similar manner to the "5-10-5" fertilizer described in this exercise, how would you describe a fertilizer in which the mass ratio of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\) to KCl is 5.00:1.00? (d) Can a "5-10-5" fertilizer be prepared in which \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\) and \(\mathrm{KCl}\) are the sole fertilizer components, with or without inert nonfertilizer additives? If so, what should be the proportions of the constituents of the fertilizer mixture? If this "5-10-5" fertilizer cannot be prepared, why not?

Liquid ethyl mercaptan, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{S},\) has a density of \(0.84 \mathrm{g} / \mathrm{mL} .\) Assuming that the combustion of this compound produces only \(\mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{SO}_{2},\) what masses of each of these three products would be produced in the combustion of \(3.15 \mathrm{mL}\) of ethyl mercaptan?

Write a formula for (a) the chloride of titanium having Ti in the O.S. \(+4 ;\) (b) the sulfate of iron having Fe in the O.S. \(+3 ;(c)\) an oxide of chlorine with Cl in the O.S. \(+7 ;\) (d) an oxoanion of sulfur in which the apparent O.S. of \(S\) is +7 and the ionic charge is \(2-\).

Two compounds of \(\mathrm{Cl}\) and \(\mathrm{X}\) are found to have molecular masses and \% Cl, by mass, as follows: 137 u, \(77.5 \% \mathrm{Cl} ; 208 \mathrm{u}, 85.1 \%\) Cl. What is element \(\mathrm{X} ?\) What is the formula for each compound?

A 0.732 g mixture of methane, \(\mathrm{CH}_{4}\), and ethane, \(\mathrm{C}_{2} \mathrm{H}_{6},\) is burned, yielding \(2.064 \mathrm{g} \mathrm{CO}_{2} .\) What is the percent composition of this mixture (a) by mass; (b) on a mole basis?
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