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Calculate empirical formula when given mass data
Calculate empirical formula when given percent composition data
Determine identity of an element from a binary formula and a percent composition
Determine identity of an element from a binary formula and mass data
Problem #1: A 15.67 g sample of a hydrate of magnesium carbonate was heated, without decomposing the carbonate, to drive off the water. The mass was reduced to 7.58 g. What is the formula of the hydrate? |
Problem #2: A hydrate of Na_{2}CO_{3} has a mass of 4.31 g before heating. After heating, the mass of the anhydrous compound is found to be 3.22 g. Determine the formula of the hydrate and then write out the name of the hydrate. |
Problem #3: When you react 3.9267 grams of Na_{2}CO_{3} · nH_{2}O with excess HCl(aq), 0.6039 grams of a gas is given off. What is the number of water molecules bonded to Na_{2}CO_{3} (value of n)? |
Problem #4: If 1.951 g BaCl_{2} · nH_{2}O yields 1.864 g of anhydrous BaSO_{4} after treatment with sulfuric acid, calculate n. |
Problem #5: Given that the molar mass of Na_{2}SO_{4} · nH_{2}O is 322.1 g/mol, calculate the value of n. |
Problem #6: Anhydrous lithium perchlorate (4.78 g) was dissolved in water and re-crystalized. Care was taken to isolate all the lithium perchlorate as its hydrate. The mass of the hydrated salt obtained was 7.21 g. What hydrate is it? |
Problem #7: A substance was found to have the following percentages by mass: 23% zinc; 11% sulfur; 22% oxygen; 44% water. What is the empirical formula? |
Problem #8: A 5.00 g sample of hydrated barium chloride, BaCl_{2} · nH_{2}O, is heated to drive off the water. After heating, 4.26 g of anhydrous barium chloride, BaCl_{2}, remains. What is the value of n in the hydrate's formula? |
Problem #9: A 1.98 g sample of a cobalt(II) chloride hydrate is heated over a burner. When cooled, the mass of the remaining dehydrated compound is found to be 1.55 g. What is the formula for the original hydrate? How can you make sure that all of the water of hydration has been removed? |
Problem #10: A solution was made by dissolving 52.0 g of hydrated sodium carbonate in water and making it up to 5.00 dm^{3} of solution. The concentration of the solution was determined to be 0.0366 M. Determine the formula of hydrated sodium carbonate. |
Problem #1: A 15.67 g sample of a hydrate of magnesium carbonate was heated, without decomposing the carbonate, to drive off the water. The mass was reduced to 7.58 g. What is the formula of the hydrate?
Solution:
1) Determine mass of water driven off:
15.67 minus 7.58 = 8.09 g of water
2) Determine moles of MgCO_{3} and water:
MgCO_{3} ⇒ 7.58 g / 84.313 g/mol = 0.0899 mol
H_{2}O ⇒ 8.09 g / 18.015 g/mol = 0.449 mol
3) Find a whole number molar ratio:
MgCO_{3} ⇒ 0.0899 mol / 0.0899 mol = 1
H_{2}O ⇒ 0.449 mol / 0.0899 mol = 5MgCO_{3} · 5H_{2}O
Problem #2: A hydrate of Na_{2}CO_{3} has a mass of 4.31 g before heating. After heating, the mass of the anhydrous compound is found to be 3.22 g. Determine the formula of the hydrate and then write out the name of the hydrate.
Solution:
1) Determine mass of water driven off:
4.31 minus 3.22 = 1.09 g of water
2) Determine moles of Na_{2}CO_{3} and water:
Na_{2}CO_{3} ⇒ 3.22 g / 105.988 g/mol = 0.0304 mol
H_{2}O ⇒ 1.09 g / 18.015 g/mol = 0.0605 mol
3) Find a whole number molar ratio:
Na_{2}CO_{3} ⇒ 0.0304 mol / 0.0304 mol = 1
H_{2}O ⇒ 0.0605 mol / 0.0304 mol = 2Na_{2}CO_{3} · 2H_{2}O
sodium carbonate dihydrate
Comment: sodium carbonate forms three hydrates and the above is not one of them. This is a problem probably crafted so that you cannot look up possible answers via the InterTubez®. Just sayin'.
Problem #3: When you react 3.9267 grams of Na_{2}CO_{3} · nH_{2}O with excess HCl(aq), 0.6039 grams of a gas is given off. What is the number of water molecules bonded to Na_{2}CO_{3} (value of n)?
Solution:
1) Some preliminary comments:
Ignore the water of hydration for a moment.Na_{2}CO_{3} + 2HCl ---> CO_{2} + 2NaCl + H_{2}O
The key is that there is a 1:1 molar ratio between Na_{2}CO_{3} and CO_{2}
2) Determine moles of CO_{2}:
0.6039 g / 44.009 g/mol = 0.013722 mol of CO_{2}
3) Use the 1:1 molar ratio referenced above:
This means that the HCl reacted with 0.013722 mole of sodium carbonate.
4) How many grams of Na_{2}CO_{3} is that?
0.013722 mol times 105.988 g/mol = 1.4544 g
5) Determine grams, then moles of water
3.9267 g minus 1.4544 g = 2.4723 g of water2.4723 g / 18.015 g/mol = 0.13724 mol of water
6) For every one Na_{2}CO_{3}, how many waters are there?
0.13724 mol / 0.013722 mol = 10Na_{2}CO_{3} · 10H_{2}O
Comment: this is one of the three sodium carbonate hydrates that exists.
Problem #4: If 1.951 g BaCl_{2} · nH_{2}O yields 1.864 g of anhydrous BaSO_{4} after treatment with sulfuric acid, calculate n.
Solution:
1) Calculate mass of Ba in BaSO_{4}:
1.864 g times (137.33 g/mol / 233.39 g/mol) = 1.0968 g
2) Calculate mass of anhydrous BaCl_{2} that contains 1.0968 g of Ba:
1.0968 g is to 137.33 g/mol as x is to 208.236 g/molx = 1.663 g
3) Calculate mass of water in original sample:
1.951 g minus 1.663 g = 0.288 g
4) Calculate moles of anhydrous BaCl_{2} and water:
1.663 g / 208.236 g/mol = 0.0080
0.288 g / 18.015 g/mol = 0.0160
5) Express the above ratio in small whole numbers with BaCl_{2} set to a value of one:
1 : 2BaCl_{2} · 2H_{2}O
Problem #5: Given that the molar mass of Na_{2}SO_{4} · nH_{2}O is 322.1 g/mol, calculate the value of n.
Solution:
1) The molar mass of anhydrous Na_{2}SO_{4} is:
142.041 g/mol
2) The mass of water in one mole of the hydrate is:
322.1 g - 142.041 g = 180.059 g
3) Determine moles of water:
180.059 g / 18.0 g/mol = 10 mol
4) Write the formula:
Na_{2}SO_{4} · 10H_{2}O
Problem #6: Anhydrous lithium perchlorate (4.78 g) was dissolved in water and re-crystalized. Care was taken to isolate all the lithium perchlorate as its hydrate. The mass of the hydrated salt obtained was 7.21 g. What hydrate is it?
Solution:
1) The amount of water in the hydrate is:
7.21 g minus 4.78 g = 2.43 g
2) The moles of anhydrous LiClO_{4} and water are:
LiClO_{4} ⇒ 4.78 g / 106.39 g/mol = 0.044929 mol
H_{2}O ⇒ 2.43 g / 18.015 g/mol = 0.13489 mol
3) Determine whole number ratio:
LiClO_{4} ⇒ 0.044929 mol / 0.044929 mol = 1
H_{2}O ⇒ 0.13489 mol / 0.044929 mol = 3LiClO_{4} · 3H_{2}O
Problem #7: A substance was found to have the following percentages by mass: 23% zinc; 11% sulfur; 22% oxygen; 44% water. What is the empirical formula?
Solution:
1) Assume 100 g of the compound is present, then find the moles of each:
Zn ⇒ 23 /65.4 = 0.352
S ⇒ 11/32 = 0.344
O ⇒ 22/16 = 1.375
H_{2}O ⇒ 44/ 18 = 2.44
2) Divide the smallest number into the others. The answers will not be exact but enough to tell the formula:
Zn ⇒ 0.352 / 0.344 = 3
S ⇒ 0.344 / 0.344 = 1
O ⇒ 1.375 / 0.344 = 3
H_{2}O ⇒ 2.44 / 0.344 = 7The formula is ZnSO_{4} · 7H_{2}O
Problem #8: A 5.00 g sample of hydrated barium chloride, BaCl_{2} · nH_{2}O, is heated to drive off the water. After heating, 4.26 g of anhydrous barium chloride, BaCl_{2}, remains. What is the value of n in the hydrate's formula?
Solution:
1) Calculate moles of anhydrous barium chloride:
4.26 g / 208.236 g/mol = 0.020458 mol
2) Calculate moles of water:
5.00 minus 4.26 = 0.74 g0.74 g / 18.015 g/mol = 0.041077 mol
3) Determine whole number ratio:
0.041077 / 0.020458 = 2
4) Formula is:
BaCl_{2} · 2H_{2}O
Problem #9: A 1.98 g sample of a cobalt(II) chloride hydrate is heated over a burner. When cooled, the mass of the remaining dehydrated compound is found to be 1.55 g. What is the formula for the original hydrate? How can you make sure that all of the water of hydration has been removed?
Solution:
1) determine mass of water driven off:
1.98 g minus 1.55 g = 0.43 g
2) Determine moles of anhydrous CoCl_{2} and H_{2}O:
CoCl_{2} ⇒ 1.55 g / 129.839 g/mol = 0.01194 mol
H_{2}O ⇒ 0.43 g / 18.015 g/mol = 0.0239 mol
3) Look for lowest whole-number ratio:
CoCl_{2} ⇒ 0.01194 mol / 0.01194 mol = 1
H_{2}O ⇒ 0.0239 mol / 0.01194 mol = 2
4) Formula is:
CoCl_{2} · 2H_{2}O
5) How can you make sure that all of the water of hydration has been removed?
After weighing the anhydrous CoCl_{2}, you would continue to heat it. Then, you would weigh it again. If the two weights are in agreement, then you are done heating. If the two weights disagree, you continue heating and weighing until you gets weights that agree. In some cases, where an extra amount of care must be taken, you would want three straight weighings that were in agreement.Also, weighs being in agreement does not mean that they are exactly the same. The standards for being in agreement might vary from one instructor to the next, so make sure to consult with your lab teacher on this point.
Problem #10a: A solution was made by dissolving 52.0 g of hydrated sodium carbonate in water and making it up to 5.00 dm^{3} of solution. The concentration of the solution was determined to be 0.0366 M. Determine the formula of hydrated sodium carbonate.
Solution:
1) moles of hydrated sodium carbonate in 5.00 liters:
0.0366 mol/L times 5.00 L = 0.183 mol
2) molecular weight of hydrated sodium carbonate:
52.0 g / 0.183 mol = 284.153 g/mol
3) mass of water in one mole of hydrate:
284.153 - 105.988 = 178.165 g(105.988 is molar mass of anhydrous sodium carbonate)
4) moles of water in one mole of hydrate:
178.165 g / 18.018 g/mol = 9.9 mol
5) formula of hydrated sodium carbonate:
Na_{2}CO_{3} · 10H_{2}O
Problem #10b: A solution was made by dissolving 71.50 g of hydrated sodium carbonate in water and making it up to 5.00 L of solution. The concentration of the solution was found to be 0.04805 M. What is the water of hydration for this hydrate of sodium carbonate?
Solution:
1) Mass of sodium carbonate in 5.00 L:
(0.04805 mol/L) (5.00 L) = x / 105.988 g/molx = 25.4636 g
2) Mass of water in 71.50 g of hydrated Na_{2}CO_{3}:
71.50 g minus 25.4636 g = 46.0364 g
3) moles of each:
moles of water ---> 46.0364 g / 18.015 g/mol = 2.55545 mol
moles of Na_{2}CO_{3} ---> 25.4636 g / 105.988 g/mol = 0.24025 mol
4) Smallest whole-number ratio:
The ratio we want in smallest whole-numbers is this ---> 0.24025 mol to 2.55545 molThat is a 1 to 10 ratio.
Na_{2}CO_{3} · 10H_{2}O
Problem #11: Determine the formula and name for the hydrate: 73.42% ammonium phosphate and 26.58% water.
Solution:
1) Assume 100 grams of the compound is present. Therefore:
73.42 g of (NH_{4})_{3}PO_{4}
26.58 g of H_{2}O
2) Determine the moles of each compound:
73.42 g / 149.0858 g/mol = 0.492468 mol
26.58 g / 10.185 g/mol = 1.475437 mol
3) We want to know how many moles of H_{2}O are present for every one mole of (NH_{4})_{3}PO_{4}:
1.475437 mol / 0.492468 mol = 2.996 = 3
4) Formula and name:
(NH_{4})_{3}PO_{4} · 3H_{2}Oammonium phosphate trihydrate.
Bonus Problem: A hydrate of magnesium chloride is present and the following data is collected:
mass of crucible = 22.130 grams
mass of crucible + hydrate = 25.290 grams
mass of crucible and contents after heating = 23.491 grams
What is the complete formula of this hydrate?
1) mass of hydrate:
25.290 g - 22.130 g = 3.160 g
2) mass of anhydrate:
23.491 g - 22.130 g = 1.181 g
3) water lost:
3.160 g - 1.181 g = 1.979 g
4) moles MgCl_{2}:
1.181 g / 95.211 g/mol = 1.24 mol
5) moles water lost:
1.979 g / 18.015 g/mol = 0.109853 mol
6) molar ratio of MgCl_{2} to water is:
1 : 11.3Within fairly reasonable experimental error, the formula of the hydrate is:
MgCl_{2} · 12H_{2}O
The Wiki page for magnesium chloride shows hydrates with 12, 8, 6, 4 and 2 waters of hydration exist. The one that exists at room temperature is the one with 6. The one with 12 loses 4 waters of hydration above -16.4 °C. So, while the problem above does not occur near room temperature, it can occur.
Here is a Yahoo Answers solution to a waters of hydration problem which I answered. Another person answered it as well and their approach differed from mine. You may want to examine it to see the differences.
Return to Mole Table of Contents
Calculate empirical formula when given mass data
Calculate empirical formula when given percent composition data
Determine identity of an element from a binary formula and a percent composition
Determine identity of an element from a binary formula and mass data