Manganese can form many different oxides such as MnO_2 or Mn_2 O_7. If an unknow

Question

Manganese can form many different oxides such as MnO_2 or Mn_2 O_7. If an unknown pure compound of the oxide of manganese was found to be 77.45% manganese, then the empirical formula of the compound is: MnO MnO Mn_2 O_7 Mn_3 O_4 A laboratory chemist made a compound that was known to contain carbon, hydrogen, and another element that was not oxygen, 52.45 mg of the compound was and 0126 grams of carbon dioxide and 0.02164 grams of water were produced. The chemist determined that the ratio of moles of carbon: moles of Hydrogen was 1:1 3:2 6:5 2:3 Vinegar is a solution that is 95% (by mass) water and 5% by mass acetic acid (C_2 H_3 O_2). IF the mass of a solution was measured to be 1230 g. then how many moles of acid are in that solution? 0.00104 moles 1.04 moles 0.065 moles 65 moles A student noticed that when a sample of a hydrocarbon of unknown formula Vs combusted to determine its empirical formula, the mass of the products the carbon dioxide and water) is more than the mass of the unknown compound. The student says "these oxide masses can't be right - the mass of the products can't be more than the mass of the reactants." The student is Wrong because the mass of products can be more than mass of reactants. Wrong because the student forgot to include the oxygen as part of the reactants. Right because of conservation of mass Right because in real life the mass of the products is always less than the mass of the reactants. (The percent yield is always less than 100%) Suppose, in a different universe, the mole was defined to be the equivalent of 3.012 Times 10^23 particles instead of Avogadro's number. How would this affect the balanced chemical equations? All the stoichiometric coefficients would be twice as big as in our universe.

Explanation / Answer

Question 11

Let's take a basis of 100g of compound.

It would have 77.45g Mn and 22.55g O.

Let's convert those grams to moles of each element using their atomic masses (Mn = 54.94 g/mol) (O = 16 g/mol).

mol Mn = 77.45/54.94 = 1.41

mol O = 22.55/16 = 1.41

Since both moles are the same, this means the formula is A) MnO.

Question 12

We calculate the moles of H present in the compound taking the moles of H2O produced as a basis and we do the same for C using the moles of CO2: We divide the grams by the molar mass of H2O (MM = 18.015 g/mol) and CO2 (MM = 44.009 g/mol). Then we convert to moles of H (by multiplying moles of water times 2 because there are two H atoms in one water molecule) and moles of C (they are the same as moles of CO2)

g H2O = 0.02164g

mol H2O = 0.001201

mol H = 0.002402

g CO2 = 0.1269g

mol CO2 = 0.002883

mol C = 0.002883

mol C / mol H = 0.002883 / 0.002402

mol C / mol H = 1.2 = 6/5

The correct answer is C) 6:5

Question 13

If we take 1230g as a basis, this means we have 61.5g of acetic acid (1230 x 0.05). The molar mass of acetic acid is MM=59.044 g/mol. We convert those grams to moles using this molar mass:

moles = g/MM

moles = 61.5g/(59.044g/mol)

moles = 1.04 moles acetic acid

The correct answer is B)1.04 moles.

Question 14

The correct answer is B. Yes, the students needs to include the mass of oxygen for the law of conservation of mass to be correct.

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