C105 Mastering HW 4&5

Butane (C4H10) is the liquid fuel in lighters. How many grams of carbon are present within a lighter containing 7.25 mL of butane? (The density of liquid butane is 0.601 g/mL.)

(0.601 g/mL) X (7.25 mL) X (48 g C / 58 g butane) = 3.61 g carbon The mass percent of carbon in butane is 82.7%. The given volume of butane (7.25 mL) multiplied by its density (in g/mL) yields the mass of butane, which is 4.36 g. The mass of carbon in the lighter is then calculated by multiplying the mass of butane by the mass percent ratio of carbon in butane (4.36 g×0.827=3.60 g).

Calculate the mass percent composition of nitrogen in (NH4)2SO4.

(NH4)2SO4 = 132.1395 g/mol N2 = 28.01344 g/mol (28.01344 g/mol) / (132.1395 g/mol) = 0.211999 = 21.1999% N in (NH4)2SO4

Part complete A flask contains 0.110 mol of liquid bromine, Br2. Determine the number of bromine molecules present in the flask.

0.11*6.022*10^23 = 6.624*10^22 molecules

An element has two naturally occurring isotopes. Isotope 1 has a mass of 120.9038 amu and a relative abundance of 57.4%, and isotope 2 has a mass of 122.9042 amu.Find the atomic mass of this element.

1. Atomic mass1 = 120.9038 x 0.574 -> 69.3988 amu 2. Atomic mass2 = 122.9042 x 0.426 -> 52.4012 amu 3. Atomic mass = 69.3988 + 52.4012 -> 121.8 amu Answer = 121.8 amu; Sb

Calculate the mass of 1.00×1024 (a septillion) molecules of water.

1.00 x10^24/6.02x10^23 = 1.661 moles 1 mol of H2O weighs 18g 1.661 * 18 = 29.9 g

What is the formula mass of Mg(NO3)2?

148.3 amu

The ion S2− has _____ protons and _____ electrons.

16, 18

Find the number of ibuprofen molecules in a tablet containing 200.0 mg of ibuprofen (C13H18O2).

200 mg to g Empirical formula mass .2 g/ empirical formula mass Avagadro's number x number 5.839 x 10^20 molecules

Specify the number of protons, neutrons, and electrons in the neutral atom copper-64

29, 35, 29

A certain metal hydroxide, M(OH)2, contains 32.8% oxygen by mass. What is the identity of the metal M?

2MO/2MH+2MO+x= 2⋅16/ 2⋅1+2⋅16+x=0.328 By simple algebraic manipulation, we get x = 63.56 g/mol Copper

A compound is 40.0% C, 6.70% H, and 53.3% O by mass. Assume that we have a 100.-g sample of this compound. What are the subscripts in the empirical formula of this compound? Enter the subscripts for C, H, and O

40 g ÷ 12 g/mol = 3.33 moles C 6.70 g ÷ 1 g/mol = 6.70 moles H 53.3 g ÷ 16 g/mol = 3.33 moles O. Dividing through by the lowest mole amount, the ratios of C : H : O is 1 : 2.01 : 1, so the emperical formula is CH2O.

Part complete Calculate the formula mass of ethanol, C2H5OH.

46.1 amu

What is the natural abundance of Br-79?

50.69% The sum of percent abundances for all isotopes is equal to 100%, so the percent abundance of Br-79 (more commonly written as 79Br) is the difference between 100% and 49.31%. Now that you have the percent abundance of the other major isotope, you can use the atomic mass to calculate the mass of this isotope.

^^ A compound is 80.0% carbon and 20.0% hydrogen by mass. Assume a 100.-g sample of this compound. How many grams of each element are in this sample?

80gC 20gH 6.67mol C 19.8molH CH3=empirical formula

Calculate the mass of Br-79.

Atomic masses are calculated as weighted averages because the distribution of isotopes is uneven. The following calculation shows how the atomic mass of bromine, which was given in the introduction, is calculated using the isotope masses: 0.5069(78.92 amu)+0.4931(80.9163 amu)=79.904 amu=79.90 amu Therefore, the isotope mass of 79Br is valid.

Main Group elements: Transition metals: Inner transition metals:

Ba,N,Pb Pd,Co Fm

Calculate the mass percent composition of nitrogen in CO(NH2)2.

CO(NH2)2 = 60.0553 g/mol N2 = 28.01344 g/mol (28.01344 g/mol) / (60.0553 g/mol) = 0.466461 = 46.6461% N in CO(NH2)2

What isotope has 17 protons and 18 neutrons?

Chlorine-35

What is the mass of a sample of water containing 3.55×1022 molecules of H2O?

Divide number of molecules by avagadros Multiply by empirical formula mass 1.06g

What is the empirical formula for the compound P4O6?

Divide ration by half P2O3

^^ The molecular formula mass of this compound is 120 amu . What are the subscripts in the actual molecular formula?

Mass of empirical formula (CH2O) = 30.03 amu n= molecular formula mass/empirical formula mass n= 120 amu/30.03 amu n=4 (CH2O)4 = C4H8O4

If the fluoride is consumed as sodium fluoride (45.24% F), what amount of sodium fluoride contains the recommended amount of fluoride?

Mass percent can be used to determine the mass of a constituent for any mass of the substance. For example, multiplying 100 mg of sodium fluoride by 0.4524 (the mass ratio in decimal form) yields that it contains 45.24 mg of fluoride. This conversion aids in determining the amount of sodium fluoride needed to get the recommended amount of fluoride using the following conceptual scheme: mgF−→mgNaF These masses can be used as a conversion factor as follows: 3.0mgF−×100.0gNaF45.24mgF−=6.6mgNaF

Which element does X represent in the following expression: 55X25?

Mn

Which fertilizer has the highest nitrogen content? Which fertilizer has the highest nitrogen content? NH3 CO(NH2)2 NH4NO3 (NH4)2SO4

NH3 mass of element ----------------------- X 100 gram-formula mass NH3 = 82.25% CO(NH2)2 = 23.33% NH4NO3 = 17.51% (NH4)2SO4 = 10.60%

A hydrocarbon is a compound that contains mostly carbon and hydrogen. Calculate the percent composition (by mass) of the following hydrocarbon: C 5 H 12 . note: Percent composition refers to the mass percent of each element in a compound: mass percent=mass of elementmass of compound×100% For example, the percent composition of water, H2O, is 11.2% hydrogen and 88.8% oxygen. Therefore, a 100-g sample of water contains 11.2 g of hydrogen atoms and 88.8 g of oxygen atoms.

mass of compound= 72.096amu mass of element C-60 H-12.098 60/72.096 *100=83.22% 12.098/72.096 *100= 16.78%

What is the symbol (including the atomic number, mass number, and element symbol) for the oxygen isotope with 10 neutrons?

18O8

An element has four naturally occurring isotopes with the masses and natural abundances given here. Isotope Mass (amu) Abundance (%) 1 135.90714 0.1900 2 137.90599 0.2500 3 139.90543 88.43 4 141.90924 11.13 Find the atomic mass of the element. Express your answer to four significant figures and include the appropriate units.

Atomic Mass = (135.90714 X 0.001900) + (137.90599 X 0.002500) + (139.90543 X 0.8843) + (141.90924 X 0.1113) = 140.1 amu Cerium

The fuel used in many disposable lighters is liquid butane, C4H10. How many carbon atoms are in 3.50 g of butane?

First convert grams to moles using molar mass of butane that is 58.1 g 3.50g C4H10 x (1 mol C4H10)/(58.1g C4H10) = 0.06024 mol C4H10 Now convert moles to molecules by using Avogadro's number 0.06024 mol C4H10 x (6.022x10^23 molecules C4H10)/(1 mol C4H10) = 3.627x10^22 molecules C4H10 And there are 4 carbon atoms in 1 molecule of butane, so use the following ratio: 3.627 x 10^22 molecules C4H10 x (4 atoms C)/(1 molecule C4H10) = 1.45 x 10^23 atoms of carbon are present

Calculate the mass percent composition of nitrogen in NH3.

NH3 = 17.03056 g/mol N = 14.00672 g/mol (14.00672 g/mol) / (17.03056 g/mol) = 0.822446 = 82.2446% N in NH3

Calculate the mass percent composition of nitrogen in NH4NO3.

NH4NO3 = 80.0434 g/mol N2 = 28.01344 g/mol (28.01344 g/mol) / (80.0434 g/mol) = 0.349978 = 34.9978% N in NH4NO3

Determine the mass of oxygen in a 7.2-g sample of Al2(SO4)3.

The mass percent of oxygen in aluminum sulfate (molar mass = 342.132 g/mol) is 56.1%, which is calculated as follows: mass percent of O in Al2(SO4)3=12×15.998 g/mol342.132 g/mol The mass of oxygen in 7.2 g of aluminum sulfate is then calculated by multiplying the mass percent ratio of oxygen by the sample mass (7.2 g×0.561=4.0 g). ORRRRRR 7.2 g / 342 g/mol = 0.0210 moles Al2(SO4)3 0.0210 moles Al2(SO4)3 X (12 mol O / 1 mol Al2(SO4)3) X 16.0 g/mol O = 4.0 g O

Ibuprofen has the following mass percent composition: C:75.69% H:8.80% O:15.51% What is the empirical formula of ibuprofen?

With all these sorts of problems, we assume a mass of 100 ⋅ g and we divide thru by the ATOMIC mass of each component: Moles of carbon = 75.69g /12.011g⋅mol−1 =6.30⋅mol Moles of hydrogen =8.80⋅g/1.00794⋅g⋅mol−1=8.73⋅mol . Moles of oxygen =15.51⋅g/15.999⋅g⋅mol−1=0.969⋅mol We divide thru by the LOWEST molar quantity (that of oxygen) to get a trial empirical formula of C6.50H9O . But, by definition, the empirical formula is the SIMPLEST WHOLE NUMBER ratio defining constituent atoms in a species. So we double this trial formula to get............ C13 H18 O2