eRification of a Compound by Mass Relationships 1. A student attempts to identif
ID: 576406 • Letter: E
Question
eRification of a Compound by Mass Relationships 1. A student attempts to identify an unknown compound by the method used in this expeni- ment. She finds that when she heated a sample weighing 0.5015 g the mass went down appreciably, to 0.3432 g. When the product was converted to a chloride, the mass went up, to 0.3726 g. a. Is the sample a carbonate? Give your reasoning. b. What are the two compounds that might be in the unknown? and Write the chemical equation for the overall reaction that would occur when the origi- nal compound was converted to a chloride. If the compound is a hydrogen carbonate, use the sum of Equations 1 and 2. If the sample is a carbonate, use Equation 2. Write the equation for a sodium salt and then for a potassium salt. c. d. How many moles of the chloride salt would be produced from one mole of original compound? e. How many grams of the chloride salt would be produced from one molar mass of original compound? g Na,Co, g NaC1 g Molar masses: NaHCO, g original compound g chloride If a sodium salt, chloride g on f. What is the theoretical mass ratio, mass chloride/mass original compound? :1 If she has a K salt, If she has an Na salt, (continued)Explanation / Answer
1.(a to d) If the sample is a metallic carbonate, a metallic oxide and carbon dioxide was produced as the sample was heated. The mass of the metallic oxide is 0.3432 grams.
Let’s determine the mass of CO2.
Mass of CO2 = 0.5015 – 0.3432 = 0.1583
The CO2 leaves the crucible.
Mass of the metallic chloride = 0.3726 grams
To produce the metallic chloride, metallic oxide could be mixed with hydrochloric acid.
MO + HCl MCl + H2O
Balance the equation.
MO + 2 HCl 2 MCl + H2O
The ratio of moles of MO to moles of MCl is 1 : 2
Moles of MO = Mass of MO ÷ molar mass = 0.3432 ÷ molar mass of MO
Moles of MCl
1 / 2 = (0.3432 ÷ molar mass of MO) / (0.3726 ÷ molar mass of MCl)
Multiply both sides by (0.3726÷ molar mass of MCl)
(0.3726 ÷ molar mass of MCl) * ½ = (0.3432 ÷ molar mass of MO)
(0.1863 ÷ molar mass of MCl) = (0.3432 ÷ molar mass of MO)
Multiply both sides by (molar mass of MO)
(molar mass of MO) * (0.1863 ÷ molar mass of MCl) = 0.3432
Divide both sides by 0.1863
(molar mass of MO) ÷ (molar mass of MCl) = 0.3432 ÷ 0.0.1863 1.842
The molar mass of the metallic oxide is approximately 1.842 times the molar mass of the metallic chloride.= Mass of MCl ÷ molar mass = 0.3726 ÷ molar mass of MCl
E.NaHCO3=84g,Na2CO3= 106g NaCl=58.5g, KHCO3=100,K2CO3=137,KCl=74.5
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.