noHz10-0.533 18 2. A hydrate co hydrate had weighed 25g. Analysis of the anhydro

Question

noHz10-0.533 18 2. A hydrate co hydrate had weighed 25g. Analysis of the anhydrous substance of S, and 6.4g of O were present. Find the formula of the hydrate. containing copper, sulfur, oxygen, and water lost 9g upon heating. Originally the revealed that the 6.4g of Cu, 3 eles cu:o-100 7 moles of S.· 3.3 :0.1 Meles o: o.4 37 CUsou5Hh0 hydrate 259 Water lost 99 5 3. When heated in the presence of oxygen, 12g of the magnesium forms a magnesium oxide compound weighing 20g. Find the empirical formula of the compound formed. ple of gas occupies 2.0 liters at STP. The sample contains 2.143g of carbon and 0.358 of hydrogen. Find the empirical and molecular formulas of the gas. 5. A compound was found to contain 55.2 percent xenon and 44.8 percent chlorine. Find the empirical formula. 26 W Chem_ Kime-Hunt (ex 4):22-26:8

Explanation / Answer

3. Atomic weight of Magnesium is 24 grams/ mole

12 grams of Mg = 12/ 24 mole = 0.5 mole magnesium

Weight of magnesium oxide = 20 grams

Weight of oxygen in the oxide is (20-12) grams = 8 grams

8 grams of oxygen = 8/16 mole = 0.5 mole oxygen

So, mole ratio of Mg and O in the oxide is 0.5:0.5 = 1:1

The simplest formula of the oxide is MgO.

This is the empirical formula of magnesium oxide.

4. The ideal gas law is: P * V = n * R * T

P = pressure of the gas = 1 atm, V = volume = 2.0 L, R = universal gas constant = 0.082 Lit*atm*mole-1 * K-1 , T = 273 K, n = number of moles of the gas

n = (P*V)/ (R*T) = (1 atm * 2.0 Lit)/ (0.082 Lit*atm*mole-1 * K-1 * 273 K) = 0.09 mole

n = w/M where w is the weight of the gas = (2.143 + 0.358) grams = 2.501 grams, M = molar mass of the gas

M = w/n = 2.501/ 0.09 grams/ mole = 27.8 grams/ mole

In the gas, the mole ratio of Carbon and Hydrogen is = (2.143/12): (0.358/1) = 0.18 : 0.358

(atomic mass of carbon and hydrogen are 12 g/ mole and 1 g/ mole respectively)

The simplest ratio of C and H in the molecule is: (0.18/0.18): (0.358/0.18) = 1:2 (approximately)

The simplest mole ratio of C and H in the gas molecule is : 1:2

The empirical formula is : CH2. Suppose , the molecular formula is (CH2)x where x is any integer

The molar mass of the gas according to empirical formula is 14 * x grams/ mole

So, 14 * x = 27.8,      Or, x = 2 (approximately)

So, the molecular formula is (CH2)2 or C2H4

5. Atomic mass of Xe is 131.29 grams/ mole and atomic mass of chlorine is 35.5 grams/ mole

Mole ratio of Xenon and chlorine in the molecule is: (55.2/131.29) : (44.8/35.5) = 0.42: 1.26

The simplest mole ratio of Xe and Cl is : (0.42/0.42): (1.26/0.42) = 1:3

So, the empirical formula is XeCl3

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