Departmental Problems for Chapter 3: Molecules, Compounds, and Chemical Equation

Question

Departmental Problems for Chapter 3: Molecules, Compounds, and Chemical Equations 1. A 1.000 g mixture of copper() oxide and copper(lIl) oxide was reduced quantitatively by hy drogen gas to give 0.8390 g of metallic copper. What is the mass of the copper) oxide in the original 1.000 g sample? 2. The metal, M, forms the sulfate M,(SO). An 0.738 g sample of this sulfate is converted to 1.511 g BaSO, What is the identity of M? 3. Lysine, an essential amino acid in the human body, contains C, H, O, and N. In one experiment, the complete combustion of 2.175 g of lysine gave 3.94 g CO, and 1.89 g of H,O. In a separate experiment, 1.873 g of lysine gave 0.436 g NH,. (a) Calculate the empirical formula of lysine. (b) The approximate molar mass of lysine is 150 g/mol. What is the molecular formula of the compound? 4. The aluminum sulfate hydrate (Al(SO) xHO) contains 8.10 percent of Al by mass. Calculate the value of x 5. An iron bar weighed 664 g. On standing in moist air for a month, exactly one-eighth of the iron tuned to rust (Fe,O,). Calculate the change in mass. Assume any rust formed remains on the iron bar. 6. A student took an unknown sample mass of a compound of Ti and Cl and put it into some water The Ti formed Tio, which was removed, dried, and found to weigh 0.777 g. AgNO, was then added to the solution until all the Cl-ions were converted to 5.575 g of AgC1. Determine the empirical formula of the unknown compound of Ti and Cl 7. A sample of M,0, weighing 11.205 g is converted to 15.380 g of MCl, where M is an unknown metal. What is the identity of M? 8. An unknown compound contains only C and N. Combustion of 0.9500 g of this unknown com- pound results in the formation of 0.1411 g of N,O, 0.1925 g of NO, and some CO, as the only products. Determine the empirical formula of the unknown compound. 9. ABS is a tough plastic composed of three smaller units called monomers: A-acrylonitrile (C,H,N), B-butadiene (C,HO, and s styrene (CsHs). a. A sample of ABS plastic contains 8.80% N by mass. It takes 0.605 g Br, to react completely with a 1.20 g sample of ABS plastic (Br, reacts in a 1:1 molar ratio with butadiene and noth ing else). What are the mass percents of acrylonitrile and butadiene in this plastic? b. What are the relative numbers of monomers A, B, and S in this plastic? 10. Natural rubidium has the average mass of 85.4678 u and is composed of isotopes sRb Rb/Rb in natural rubidium is 2.591. Calcu- (mass 84.9117 u) and " Rb. The ratio of atoms late the mass of Rb Numerical solutions: (1) 0.449 g Cu20; (2) Al; (3) C6H,N,02; (4) 18; (5) 36 g; (6) TiCl,i (7) Au; (8) CN; (9) 33.3% acrylonitrile and 17.1% butadiene; A,B,S,i (10) 869085

Explanation / Answer

Ans 1

Mass of Mixture of CuO and Cu2O = 1 g

Mass of Cu = 0.8390 g

Let mass of CuO = x g

Moles of CuO = mass/molecular weight = x/79.5

1 mol CuO contains = 1 mol Cu

Moles of Cu = x/79.5

mass of Cu = moles * molecular weight

= (x/79.5) * 63.5

= 0.7987x

= 63.5x/79.5 = 0.7987x

mass of Cu2O = (1 - x) g.......eq1

moles of Cu2O = mass/molecular weight

= (1-x) / 143.09

1 mol Cu2O contains = 2 mol Cu

moles of Cu = 2*(1-x) / 143

= (1-x) / 71.5

mass of Cu = moles x molecular weight

= [(1-x) / 71.5] * 63.5

= 0.8881*(1-x) g .........eq2

From Eq1 and Eq2

0.7987x + 0.8881*(1-x) = 0.8390

0.7987x + 0.8881 - (0.8881*x) = 0.8390

(0.7987x) - (0.8881x) = ( 0.8390 - 0.8881)

x = 0.549

mass of CuO = x = 0.549 g = mass of copper (I) oxide

mass of Cu2O = 1-x = 1 - 0.549 = 0.451 g

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.