
Chemical Molar Mass and Formulas

Interactive Video
•
Chemistry, Science, Mathematics
•
9th - 12th Grade
•
Hard

Patricia Brown
FREE Resource
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10 questions
Show all answers
1.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the primary difference between a molecular formula and an empirical formula?
Molecular formula shows the simplest ratio of elements.
Empirical formula shows the exact number of atoms of each element.
Empirical formula is used only for ionic compounds.
Molecular formula shows the exact number of atoms of each element.
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Why is it important to convert mass information into numerical data in chemistry?
To calculate the boiling point of a compound.
To predict the taste of a compound.
To understand the relative number of atoms in a compound.
To determine the color of compounds.
3.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
In the sample problem, why might it be beneficial to use the given mass information instead of assuming 100 grams?
It simplifies the calculation process.
It allows for more accurate results in complex problems.
It is a standard practice in all chemistry problems.
It reduces the number of calculations needed.
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the significance of distinguishing between grams and moles in chemical calculations?
Moles are only used in organic chemistry.
Grams are more accurate than moles.
Moles provide a count of particles, while grams measure mass.
Grams are used for liquids, moles for gases.
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Which molar mass is used for oxygen in the conversion from mass to moles?
12.01
1.008
14.01
16
6.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the purpose of dividing by the smallest number in empirical formula calculations?
To simplify the chemical equation.
To eliminate decimal points.
To ensure the formula is balanced.
To achieve the smallest whole number ratio.
7.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the empirical formula derived from the given problem?
C1H3O1
CH3O
C2H6O2
C3H6O3
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