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Chapter 3: Problem 150

A binary compound between an unknown element \(\mathrm{E}\) and hydrogen contains 91.27\(\% \mathrm{E}\) and 8.73\(\% \mathrm{H}\) by mass. If the formula of the compound is \(\mathrm{E}_{3} \mathrm{H}_{8},\) calculate the atomic mass of \(\mathrm{E}\)

Short Answer

Expert verified
The atomic mass of Element E is approximately 27.87 amu.

Step by step solution

01

Calculate the total mass of Hydrogen in the compound

Since the formula of the compound is E3H8, it means there are 8 hydrogen atoms in the compound. We know that the atomic mass of hydrogen is approximately 1 amu (atomic mass unit). Therefore, the total mass of hydrogen in the compound is: Total mass of hydrogen = (8 atoms of hydrogen) * (1 amu/atom of hydrogen) = 8 amu
02

Calculate the total mass of the compound

We are given the percentage composition of the compound. Since hydrogen represents 8.73% of the compound, the total mass of the compound can be expressed as: Total mass of compound = (Total mass of hydrogen) / 0.0873 Substituting the value of the total mass of hydrogen from step 1: Total mass of compound = 8 amu / 0.0873 ≈ 91.62 amu
03

Calculate the total mass of Element E in the compound

We are given that element E constitutes 91.27% of the compound. Therefore, the total mass of E in the compound can be found by using the percentage composition: Total mass of E = (Total mass of compound) * 0.9127 Substituting the value of the total mass of the compound from step 2: Total mass of E = 91.62 amu * 0.9127 ≈ 83.62 amu
04

Calculate the atomic mass of Element E

The formula of the compound, E3H8, indicates that there are 3 atoms of Element E in the compound. To find the atomic mass of E, we simply divide the total mass of E in the compound by the number of E atoms: Atomic mass of E = (Total mass of E) / 3 Substituting the value of the total mass of E from step 3: Atomic mass of E = 83.62 amu / 3 ≈ 27.87 amu Thus, the atomic mass of Element E is approximately 27.87 amu.

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Most popular questions from this chapter

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 \(\mathrm{g}\) of the compound produced 0.213 \(\mathrm{g} \mathrm{CO}_{2}\) and 0.0310 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\) . In another experiment, it is found that 0.103 \(\mathrm{g}\) of the compound produces 0.0230 $\mathrm{g} \mathrm{NH}_{3} .$ What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\) . Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\) . Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

Balance the following equations: a. $\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)$ b. $\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ c. $\mathrm{AgNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+\mathrm{HNO}_{3}(a q)$

Balance the following equations representing combustion reactions: c. $C_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ d. Fe \((s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) e. $\mathrm{FeO}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$

The reusable booster rockets of the U.S. space shuttle employ a mixture of aluminum and ammonium perchlorate for fuel. A possible equation for this reaction is $$ 3 \mathrm{Al}(s)+3 \mathrm{NH}_{4} \mathrm{ClO}_{4}(s) \longrightarrow $$ $$ \mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{AlCl}_{3}(s)+3 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ What mass of \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) should be used in the fuel mixture for every kilogram of Al?

With the advent of techniques such as scanning tunneling microscopy, it is now possible to "write" with individual atoms by manipulating and arranging atoms on an atomic surface. a. If an image is prepared by manipulating iron atoms and their total mass is \(1.05 \times 10^{-20} \mathrm{g},\) what number of iron atoms were used? b. If the image is prepared on a platinum surface that is exactly 20 platinum atoms high and 14 platinum atoms wide, what is the mass (grams) of the atomic surface? c. If the atomic surface were changed to ruthenium atoms and the same surface mass as determined in part b is used, what number of ruthenium atoms is needed to construct the surface?
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