In 1987 the first substance to act as a superconductor at a temperature above that of liquid nitrogen \((77 \mathrm{~K})\) was discovered. The approximate formula of this substance is \(\mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{7} .\) Calculate the percent composition by mass of this material.

Short Answer

Expert verified
The percent composition by mass of YBa\(_{2}\)Cu\(_{3}\)O\(_{7}\) is approximately: Y: \(\frac{88.91}{666.62}\) × 100% \(= 13.34 \% \) Ba: \(\frac{274.66}{666.62}\) × 100% \(= 41.18 \% \) Cu: \(\frac{190.65}{666.62}\) × 100% \(= 28.60 \% \) O: \(\frac{112.00}{666.62}\) × 100% \(= 16.80 \% \)

Step by step solution

01

Calculate the molar mass of the substance

First, we need to find the molar mass of the given substance (YBa\(_{2}\)Cu\(_{3}\)O\(_{7}\)). We can do this by using the periodic table and adding the molar masses of each element in the formula. \[ \begin{aligned} \mathrm{Molar\ Mass} &= 1(\mathrm{Molar\ Mass\ of\ Y}) + 2(\mathrm{Molar\ Mass\ of\ Ba}) + 3(\mathrm{Molar\ Mass\ of\ Cu}) + 7(\mathrm{Molar\ Mass\ of\ O}) \\ &= 1(88.91 \mathrm{~g/mol}) + 2(137.33 \mathrm{~g/mol}) + 3(63.55 \mathrm{~g/mol}) + 7(16.00 \mathrm{~g/mol}) \end{aligned} \]
02

Calculate the individual masses of the elements

Now, let's calculate the individual masses of the elements in the substance's formula: \[ \begin{aligned} \mathrm{Mass\ of\ Y} &= 1(88.91 \mathrm{~g/mol}) \\ \mathrm{Mass\ of\ Ba} &= 2(137.33 \mathrm{~g/mol}) \\ \mathrm{Mass\ of\ Cu} &= 3(63.55 \mathrm{~g/mol}) \\ \mathrm{Mass\ of\ O} &= 7(16.00 \mathrm{~g/mol}) \end{aligned} \]
03

Calculate the percent composition by mass

Finally, we can determine the percent composition by mass for each element in the substance. For this, we need to divide the individual mass of each element by the molar mass of the substance and multiply the result by 100. \[ \begin{aligned} \mathrm{Percent\ Composition\ of\ Y} & = \frac{\mathrm{Mass\ of\ Y}}{\mathrm{Molar\ Mass}} \times 100\% \\ \mathrm{Percent\ Composition\ of\ Ba} & = \frac{\mathrm{Mass\ of\ Ba}}{\mathrm{Molar\ Mass}} \times 100\% \\ \mathrm{Percent\ Composition\ of\ Cu} & = \frac{\mathrm{Mass\ of\ Cu}}{\mathrm{Molar\ Mass}} \times 100\% \\ \mathrm{Percent\ Composition\ of\ O} & = \frac{\mathrm{Mass\ of\ O}}{\mathrm{Molar\ Mass}} \times 100\% \end{aligned} \] Calculate the values for each element to find their percent compositions by mass.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the production of printed circuit boards for the electronics industry, a \(0.60-\mathrm{mm}\) layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)\) A plant needs to manufacture 10,000 printed circuit boards, each \(8.0 \times 16.0 \mathrm{~cm}\) in area. An average of \(80 . \%\) of the copper is remoyed from each board (density of copper \(=8.96 \mathrm{~g} / \mathrm{cm}^{3}\) ). What masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) are needed to do this? Assume \(100 \%\) yield.

What number of atoms of nitrogen are present in \(5.00 \mathrm{~g}\) of each of the following? a. glycine, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}_{2} \mathrm{~N}\) b. magnesium nitride c. calcium nitrate d. dinitrogen tetroxide

Glass is a mixture of several compounds, but a major constituent of most glass is calcium silicate, \(\mathrm{CaSiO}_{3}\). Glass can be etched by treatment with hydrofluoric acid; HF attacks the calcium silicate of the glass, producing gaseous and water-soluble products (which can be removed by washing the glass). For example, the volumetric glassware in chemistry laboratories is often graduated by using this process. Balance the following equation for the reaction of hydrofluoric acid with calcium silicate. $$ \mathrm{CaSiO}_{3}(s)+\mathrm{HF}(a q) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$

Balance the following equations: a. \(\mathrm{Cr}(s)+\mathrm{S}_{\mathrm{s}}(s) \rightarrow \mathrm{Cr}_{2} \mathrm{~S}_{3}(s)\) b. \(\mathrm{NaHCO}_{3}(s) \stackrel{\text { Heat }}{\longrightarrow} \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{KClO}_{3}(s) \stackrel{\text { Heat }}{\longrightarrow} \mathrm{KCl}(s)+\mathrm{O}_{2}(g)\) d. \(\operatorname{Eu}(s)+\mathrm{HF}(g) \rightarrow \operatorname{EuF}_{3}(s)+\mathrm{H}_{2}(g)\)

Vitamin \(\mathrm{B}_{12}\), cyanocobalamin, is essential for human nutrition. It is concentrated in animal tissue but not in higher plants. Although nutritional requirements for the vitamin are quite low. people who abstain completely from animal products may develop a deficiency anemia. Cyanocobalamin is the form used in vitamin supplements. It contains \(4.34 \%\) cobalt by mass. Calculate the molar mass of cyanocobalamin, assuming that there is one atom of cobalt in every molecule of cyanocobalamin.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free