quimica TEST #1 CHAPTER 3

Question 1

The formulas mass of a substance is the:

Covalent substances exist as :

The formula mass of a covalent substance may be correctly referred to as a

Answer 1

the sum of the average atomic masses of all the atoms in the substance’s formula.

as discrete molecules.

as a molecular mass.

Question 2

que es molecular mass

Answer 2

sumar los atomic masses de cada elemento en el compound

Question 3

formula mass for ionic compounds

Ionic substances are composed of :

Ionic compounds do NOT exist as

The formula mass for an ionic compound may NOT correctly be referred to:

The average atomic masses of the ions can be approximated to be the same as the :

diff between ioinc compound and covalent substance

Answer 3

of discrete cations and anions combined in ratios to yield electrically neutral bulk matter.

as molecules.

as a molecular mass.

the average atomic masses of the neutral atoms.

Question 4

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc.

The mole is defined as the:

The mole provides a link between the:

Answer 4

the amount of a substance containing the same number of discrete entities (such as atoms, molecules, or ions) as the number of atoms in a sample of pure carbon-12 weighing exactly 12 g.

mass of a sample and the number of atoms, molecules, or ions in that sample.

Question 5

The number of entities composing a mole has been determine to be:

This constant is named after Italian scientist _____________ and is known as ______________.

Avogadro’s Number (NA) =

The masses of 1 mole of different elements, however, are different, since the:

Answer 5

6.02214179 x 10^23.

This constant is named after Italian scientist Amedeo Avogradro and is known as Avogadro’s Number.

Avogadro’s Number (NA) = 6.022 x 1023

since the masses of the individual atoms are drastically different.

Question 6

The molar mass of an element (or compound) is :

The molar mass of any substance is numerically equivalent to its:

Example:
A single 12C atom has a mass of 12 amu.
A mole of 12C atoms have a mass of 12 g.

Answer 6

the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol).

atomic or formula mass in amu.

molar mass es g/mol and formula mass en amu

Question 7

calculations

The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds.

Question 8

to find moles

Answer 8

mass in grams divided by molar mass

Question 9

to find mass

Answer 9

moles x molar mass

Question 10

how to find number of atoms

Answer 10

mass divided by molar mas = moles

moles x avogrado’s number = number of atoms

Question 11

Given that saccharin has a molar mass of 183.18 g/mol, how many saccharin molecules are in a 40.0-mg (0.0400-g) sample of saccharin?
How many carbon atoms are in the same sample?

Question 12

Percent composition -

Example: A 10.0 g sample of a compound is determined to contain 2.5 g hydrogen and 7.5 g carbon.

Answer 12

The percentage by mass of each element in a compound.

Question 13

For compounds of known formula, the percent composition can also be derived from:

Answer 13

the formula mass and the atomic masses of the constituent elements.

ver vid de sacar percent composition from formula mass

Question 14

A compound’s empirical formula can be determined from :

1)

2)

3)

Answer 14

from the masses of its constituent elements.

1)Convert element masses to moles using molar masses.

2) Divide each number of moles by the smallest number of moles.

3) If necessary, multiply by an integer, to give the smallest whole-number ratio of subscripts.

Question 15

A compound’s empirical formula can be determined from:

1)

2)

3)

4)

Answer 15

from its percent composition.

1) Convert percent composition to masses of elements by assuming a 100 g sample of compound.

2) Convert element masses to moles using molar masses.

3) Divide each number of moles by the smallest number of moles.

4) If necessary, multiply by an integer, to give the smallest whole-number ratio of subscripts.

Question 16

The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O. What is the empirical formula for this gas?

Answer 16

27.29 g C and 72.71 g O. divido por molar mass para tener moles. el moles mas pequeño lo divido por todo y multiplicar pa q me de un complete integer.

Question 17

derivation of molecular formulas

A compound’s molecular formula can be determined from :

formula molecules =

The molecular formula is then obtained by :

Answer 17

its empirical formula and its molecular or molar mass.

molar mass/ empirical formula mass = n formula molecules

multiplying each subscript in the empirical formula by n.

Question 18

A compound has an empirical formula of CH2O (empirical formula mass = 30 amu) and a molecular mass of 180 amu.
find molecular formula

Answer 18

180 amu/30 amu = 6 formula units/molecules

(CH20) x 6 = C6H1206

Question 19

mixtures are :

They are more commonly encountered in:

Similar to a pure substance, the relative composition of a mixture plays an important role in determining :

Answer 19

samples of matter containing two or more substances physically combined

nature than are pure substances.

its properties

Question 20

Solutions occur frequently in nature.

Solutions are another term used for a :

concentration is

Answer 20

homogeneous mixture – uniform composition and properties throughout its entire volume.

the relative amount of a given solution component

Question 21

A solution consists of two components:

1)

2)

A solution in which water is the solvent is called an :

Qualitative terms used to describe a solution:

1-

2-

Answer 21

1) Solvent - component with a concentration that is significantly greater than that of all other components.

2) Solute – component that is typically present at a much lower concentration than the solvent.

an aqueous solution.

Qualitative terms used to describe a solution:

1- Concentrated – Relatively high concentration of solute.

2- Dilute – Relatively low concentration of solute.

Question 22

Molarity (M) is :

Molarity (M) is a useful concentration unit for many applications in chemistry.

formula of M

Answer 22

the number of moles of solute in exactly 1 liter (1 L) of the solution

M = mole solute /L solution

Question 23

Dilution is the process whereby :

Dilution is a common means of :

By adding _________ to a measured portion of a more concentrated stock solution, we can achieve a ___________

Answer 23

the concentration of a solution is lessened by the addition of solvent.

preparing solutions of a desired concentration.

By adding solvent to a measured portion of a more concentrated stock solution, we can achieve a particular concentration.

Question 24

The molar amount of solute (n) in a solution is equal to:

Expressions like these may be written for a solution before (1) and after (2) it is diluted:

Since the dilution process does not change the amount of solute in the solution:

Thus, these two equations may be set equal to one another to derive the dilution equation:

Other units of concentration (C) and volume (V) may be used:

Answer 24

moles of solute (n) = M x L solution

n 1 = M 1 x L 1
n 2 = M 2 x L 2

n1 = n2

M 1 x L 1 = M 2 x L 2

C 1 x V 1 = C2 x V2

Question 25

Molarity is a very useful unit for :

Answer 25

evaluating the concentration of solutions.

Question 26

other units of concentration used in various applications.

The mass percentage of a solution component is defined as the:
formula:

volume percentage is:
formula:
——————————————–
A mass-volume percent is a :

Very low solute concentrations are often expressed using appropriately small units such as:

The mass-based definitions of ppm and ppb:
formulas:

Answer 26

ratio of the component’s mass to the solution’s mass, expressed as a percentage:
mass percentage = mass of component/mass of solution x 100
——————————————
the concentration of a solution formed by dissolving a liquid solute in a liquid solvent
volume percentage = volume of solute / volume of solvent x 100
——————————————-
ratio of a solute’s mass to the solution’s volume expressed as a percentage.
——————————————–
parts per million (ppm) or parts per billion (ppb).

ppm = mass solute/ mass of solution x 10^6
ppb = mass solute/ mass of solution x 10^9