Question

Chem200 Haggard Homework Problems Composition of Substances and Solutions 1. When 28.Ograms of CO is analyzed, it is found to contain 12.0g of carbon and 16.0g of oxveen. How much carbon and how much oxygen would be found in 100.0g of CO2 2. A sample of a compound is found to contain 40.0g of calcium, 71.0g of chlorine, 4.0g of hydrogen, and 32.0g of oxygen. How much of each element would be needed to make 750.0g of this compound? 3. It is found a certain sample of a compound contains 23.0g of sodium and 35.5g of chlorine. a. What is the percent composition of this compound? b. How much chlorine would be needed to make 7.0g of this compound? 4. Determine the empirical formula for the following compounds a. A poisonous gas, cyanogen, is 46.14% carbon and 53.8% nitrogen. What is the empirical formula of cyanogen? b. Afuel is 85.59% carbon and 14.41% hydrogen. Determine the empirical formula, 5. Determine the empirical and molecular formula for the following compounds. a. A pure compound is found to have the following percent composition: 40.44% C, 7.92% H, 35.92% O, and 15.72%N. Determine the empirical formula of this compound. b. A certain compound is 5.9% hydrogen and 94.1% oxygen and has a molar mass of 34.0g/mole. Determine the empirical and molecular formula. C. A sample of a compound is found to contain 23.35g of C, 1.05g of H, 14.78g of N and 33.83g of O. The molar mass of the compound is 213.0g/mole. Determine the empirical and molecular formula of this compound. d. A certain red dye is found to be 59.988% C, 4.146% H, 12.345% N, 9.420% S, and 14.101% O. The molar mass of the dye is 680.0g/mole. Determine the empirical and molecular formula of this compound. 6. A sulfide of gold is composed of 6.31g sulfur and 25.93g of gold. Determine the percent of both elements in the compound. 7. Peaches are 8.6% sugar by mass. If a serving of peaches is 170.0grams, how many grams of sugar are in one serving? 8. A solution is made my mixing 90.0mL of alcohol with enough water to give 250.0mL of solution. What is the percent volume of alcohol in solution? 9. An alcoholic beverage is labeled 90 proof, which means the alcohol concentration is 45% by volume. How many milliliters of pure alcohol would be present in 30mL of the beverage? 10.A 150.0mL sample of salt water is evaporated to dryness. A residue of salt weighing 27.9g is left behind. Calculate the percent mass/volume of the original salt water sample.

Answer 1

We know hecemposikon r 28g co. u28g nih how Со але Pmeants units 28a co loo 3.5 has 28 and .00 12-0a has So, and 3.571 unils 2 So 42.5 and 57-143aO i omsqwetion Findoight ire Saple wnib Similar to Suuple Find how Mhe the Sanple each elemaut Multiply the unit no 32-0 weight gCagile 148 5.102 uni 150 anple in 5D 4 uni ts no Mas o each elemant neded 204.089 5.I02x Calcium 362. 2429 Chloriwe 5.102x 1 Hysragen (5-102x 4.) OnwgenC6.102x 32-0) 20. 408 9 I63-2649

3. The frmul f poruntopasition 6' element mass Sample TOtalmaus C4) Peanttoposition r Sod ium (23.Dt 35.5 39. 31 L 35 59 UY chlorine = (23-0 35.5)9 -6o. 684 and bo.684% 0 thus penant tompesiton 34.31% N the method in 1) audl2) this Cb loe Solve Can perunt Lompoitiow Chloe Go 684 CL he using ne Using bo. 6848 Smple as 7-0Sample thi meanti Sou has 60 684

TO CALCULATE EMPIRICAL FORMULA, THE FOLLOWING STEPS ARE FOLLOWED:

• Divide the given masses of elements by their gram atomic mass to get no. of moles of the element
• Express the no. of moles as a ratio. Divide each term by the lowest term in the ratio and round off to whole number/convert each term in the ratio to a whole number by multiplying by a suitable number. This ratio will give the number corresponding to the atom in the empirical formula.

Here, we have percentage composition. That is, the mass of the element in 100g of substance is given:

4a

46.14g C and 53.8g N

Gram atomic mass of C = 12g/mole , N = 14g/mole

Step 1: dividing by gram atomic mass :

C : 46.14/12 = 3.845    N: 53.8/14 = 3.843       

Step 2: Ratio (C:N) = 3.845: 3.843

Dividing by lower number (3.843) and rounding off: 1:1

Thus, 1 C : 1N is the ratio. Empirical formula = CN

4b

85.59g C and 14.41g H

Gram atomic mass of C = 12g/mole , H = 1g/mole

Step 1: dividing by gram atomic mass :

C : 85.59/12 = 7.13      H: 14.41/1 = 14.41        

Step 2: Ratio (C:H) = 7.13: 14.41        

Dividing by lower number (7.13) and rounding off: 1:2

Thus, 1 C : 2H is the ratio. Empirical formula = CH₂