Empirical formula of a compound if a 50.0 g sample of it contains 9.1g Na ,20.6 g Cr ,and 22.2 g O
1 answer:
Answer:
The empirical formula of a compound is Na2Cr2O7
Explanation:
Empirical formula is defined as the lowest ratio of elements present in any substance. In order to determine the empirical formula for a substance containing these elements we have to follow some steps.
1.In the first step, we have to divide the given mass of an element with its molecular mass to convert the mass to moles.
Similarly, the moles of Cr and O in that compound were calculated and found as 0.396 and 1.3875, respectively.
2. Then the next step is to determine the lowest moles and divide all the moles with that lowest mole. As here the lowest mole is 0.396. So on dividing the moles of Na, Cr and O with 0.396, we will obtain the ratio as 1:1:3.5.
3. As the ratio consists of a fractional number, we have to multiply it with the smallest possible integer to convert it to whole number. In this case, if we multiply it with 2 , we will get the ratio as 2:2:7.
Thus, the empirical formula for the compound can be written as Na2Cr2O7.
You might be interested in
Answer:
The given statement is true.
Enzymes which are present in the digestive tract such as salivary amylase, pepsin, trypsin, et cetera mainly catalyze the hydrolysis reaction.
The hydrolysis reaction is the reaction by which large molecules are broken down into smaller molecules with the help of water.
Most of the complex molecules or nutrients such as starch, protein et cetera are broken down into their respective smaller units with the help of hydrolysis reaction.
For example, lactase catalyzes the hydrolysis of lactose into glucose and galactose.
Answer:
191.11 grams of oxygen gas should be produced.
Explanation:
The balanced reaction is:
2 Al₂O₃ → 4 Al + 3 O₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Al₂O₃: 2 moles
- Al: 4 moles
- O₂: 3 moles
Being the molar mass of each compound:
- Al₂O₃: 102 g/mole
- Al: 27 g/mole
- O₂: 32 g/mole
By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al₂O₃: 2 moles* 102 g/mole= 204 grams
- Al: 4 moles* 27 g/mole= 108 grams
- O₂: 3 moles* 32 g/mole= 96 grams
Then you can apply the following rule of three: if by stoichiometry 108 grams of aluminum are produced along with 96 grams of oxygen, 215 grams of aluminum are produced along with how much mass of oxygen?
mass of oxygen= 191.11 grams
<u><em>191.11 grams of oxygen gas should be produced.</em></u>