Concept explainers
How many hydrogen atoms are in each of the following?
Interpretation:
The number of hydrogen atoms is to be determined in the given molecules.
Concept Introduction:
The characteristic mass of a compound is considered as the formula mass of the compound, which is obtained by adding the atomic masses of all the atoms.
A conversion factor is considered a numerical ratio in order to change one unit into another.
A chemical formula represents the precise number of atoms in a molecule, but not their structural arrangement.
Answer to Problem 42E
Solution:
The number of hydrogen atoms present in the molecules is given as follows:
Explanation of Solution
a)
The equivalence is obtained between
In order to obtain atoms of hydrogen, the conversion factor proceeds as follows:
Therefore, the number of hydrogen atoms is
b)
The equivalence is obtained between
In order to obtain atoms of hydrogen, the conversion factor proceeds as follows:
Therefore, the number of hydrogen atoms is
c)
The equivalence is obtained between
In order to obtain atoms of hydrogen, the conversion factor proceeds as follows:
Therefore, the number of hydrogen atoms is
d)
The equivalence is obtained between
In order to obtain atoms of hydrogen, the conversion factor proceeds as follows:
Therefore, the number of hydrogen atoms is
Want to see more full solutions like this?
Chapter 4 Solutions
Chemistry In Focus
- Natural rubidium has the average mass of 85.4678 u and is composed of isotopes 85Rb (mass = 84.9117 u) and 87Rb. The ratio of atoms 85Rb/ 87Rb in natural rubidium is 2.591. Calculate the mass of 87Rb.arrow_forwardNitric acid is composed of HNO3 molecules. A sample weighing 4.50 g contains 4.30 1022 HNO3 molecules. How many nitrogen atoms are in this sample? How many oxygen atoms are in 61.0 g of nitric acid?arrow_forwardAn element X bas five major isotopes, which are listed below along with their abundances. What is the element? Isotope Percent Natural Abundance Mass (u) 46x 8.00% 45.95232 47x 7.30% 46.951764 48x 73.80% 47.947947 49x 5.50% 48.947841 50x 5.40% 49.944792arrow_forward
- You find a compound composed only of element X and chlorine. and you know that the compound is 13.10% X by mass. Each molecule of the compound contains six times as many chlorine atoms as X atoms. What is element X?arrow_forwardDiamond is one form of elemental carbon. An engagement ring contains a diamond weighing 1.25 carats (1 carat = 200 mg). How many atoms are present in the diamond?arrow_forwardUse the periodic table shown in Fig. 4.9 to determine the atomic mass (per mole) or molar mass of each of the substances in column 1, and find that mass in column 2. l> Column 1 Column 2 (1) molybdenum (a) 33.99 g (2) lanthanum (b) 79.9 g (3) carbon tetrabromide (c) 95.94 g (4) mercury(II)oxide (d) l25.4g (5) titanium(iV) oxide (e) 138.9 g (6) manganese(ll) chloride (f) 143.1 g (7) phosphine, (g) 156.7 g (8) tin(II) fluoride (h) 216.6 g (9) lead(II) sulfide (i) 239.3 g (10) copper(I)oxide (j) 3316garrow_forward
- Use the average atomic masses given inside the front cover of this book to calculate the number of moles of the element present in each of the following samples. l type='a'> 4.95 g of neon 72.5 gof nickel 115 mgofsilver 6.22 g of uranium ( is a standard abbreviation meaning “micro”) 135 gof iodinearrow_forwardA 4.19-g sample of nitrous oxide (an anesthetic, sometimes called laughing gas) contains 5.73 1022 N2O molecules. How many nitrogen atoms are in this sample? How many nitrogen atoms are in 2.67 g of nitrous oxide?arrow_forwardhe chemical formula for aspirin is C9H8O4 . What is the mass percent for each element in 1 mole of aspirin? (Give your answer to four significant figures.) l type="a"> carbon > % l type="a"> hydrogen > % l type="a"> oxygen > %arrow_forward
- Using the average atomic masses given inside the front cover of this book, calculate the number of atoms present in each of the following samples. l type='a'> 1.50 g of silver, Ag 0.0015 moIe of copper, Cu 0.0015 g of copper, Cu 2.00 kg of magnesium, Mg 2.34 oz of calcium, Ca 2.34 g of calcium, Ca 2.34 moles of calcium, Caarrow_forwardIf one can find the ratio of the number of moles of the elements in a compound to one another, one can find the formula of the compound. In a certain compound of copper and oxygen, CuxOy, we find that a sample weighing 0.9573g contains 0.7649gCu. a. How many moles of Cu are there in the sample? (MolesofCu=massofCumolarmassofCu) _moles b. How many gram of O are there in the sample? The mass of the sample equals the mass of the Cu plus the mass of O. _g c. How many moles of O are there in the sample. _moles d. What is the mole ratio (molesofCu/molesofO) in the sample? _:1 e. What is the formula of the oxide? The atom ratio equals the mole ratio and is expressed using the smallest integers possible. _ f. What is the molar mass of the copper oxide? _g/molarrow_forwardComplete the following table. l> Mass of Sample Moles of Sample Atoms in Sample 5.00 g AI td> 0.00250 mol Fe td> msp;2.61024 atoms Cu 0.00250 g Mg td> msp;2.7103 mol Na msp;1.00104 atoms Uarrow_forward
- Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningIntroductory Chemistry: A FoundationChemistryISBN:9781337399425Author:Steven S. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry: Matter and ChangeChemistryISBN:9780078746376Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl WistromPublisher:Glencoe/McGraw-Hill School Pub Co