Principles of Chemistry, Chapt . 2: Atomic Structure and The Elements The Structure of Atoms
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Transcript of Principles of Chemistry, Chapt . 2: Atomic Structure and The Elements The Structure of Atoms
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Principles of Chemistry, Chapt. 2: Atomic Structure and The Elements
I.The Structure of Atomsprotons, neutrons, and electrons
II.Atomic Structure and Properties—the Elements
atomic mass, atomic number, isotopes
III.The Mole Concept: 6.02 x 1023
IV.The Periodic Table
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Homework: Chapt. 2 Problems 26, 29, 37, 43, 75
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~ 10-10 meters = 1 angstrom (Å)
_
+10-14 m
Smeared out electron charge cloud
+++
++
++
Protons and neutrons
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Most of the mass is here
Most of the Chemistry is here
Atomic Theory in a single Slide
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STM Image: Oxygen atoms at the surface of Al2O3/Ni3Al(111)
S. Addepalli, et al. Surf. Sci. 442 (1999) 3464
Electronic charge cloud surrounding the nucleus
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What’s inside the nucleus:
Particle Mass (amu) Charge
Proton (p+) 1.007 amu +1Neutron (n0) 1.009 amu 0
What’s outside the nucleus:
Electron (e-) .00055 amu -1
Note: mass ratio of electron/proton (Mp+/Me-) = 1836
For any atom: # of electrons = # of protons: Why?
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Atomic Theory: Late 19th Century
Atomic theory—everything is made of atoms—generally accepted (thanks to Ludwig Boltzman, and others).
Mendeleev/periodic table—accepted, but the basis for periodic behavior not understood
What are atoms made of?
How are they held together?
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Electrical behavior: “+” attracts “-” but like charges repel
Atoms must contain smaller sub-units.
Radioactive material
Beam of , , and Electrically
charged plates
β-particles (“–”)
Gamma ray (γ) No charge, no deflection
α-particle (“+” ) Heavier, deflected less than β
–+
Alpha particle 2 n0 + 2p+
Beta particel electron (e-)Gamma photon
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Electric and magnetic fields deflect the beam.• Gives mass/charge of e- = −5.60 x 10-9 g/C• Coulomb (C) = SI unit of charge
•Thomson (1897) discovered the e-:
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“Cathode rays”• Travel from cathode (-) to anode (+).• Negative charge (e−).• Emitted by cathode metal atoms.
fluorescentscreen
– high voltage + cathode ray
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+
_
+
_
Essence of the Thompson Experiement (and old fashioned TV’s)
Electric field exerts Force+ plate repels +charged particles- Plate repels – charged particles
F = Eq = ma
d = displacement = ½ at2 = Eq/m (t = L/Vx)
Therefore, the greater the displacement, the lower the mass of the particle
d
x
y
Phosphor screen
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• Millikan (1911) studied electrically-charged oil drops.
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Charge on each droplet was: n (−1.60 x 10-19 C) with n = 1, 2, 3,… n (e- charge)
Modern value = −1.60217653 x 10-19 C. = −1 “atomic units”.
•These experiments give:
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Modern value = 9.1093826 x 10-28 g
= (-1.60 x 10-19 C)(-5.60 x 10-9 g/C) = 8.96 x 10-28 g
me = charge xmass
charge
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• Atoms gain a positive charge when e- are lost.
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Implies a positive fundamental particle.
Hydrogen ions had the lowest mass.• Hydrogen nuclei assumed to have “unit
mass”• Called protonsprotons.Modern science: mp = 1.67262129 x 10-24 g
mp ≈ 1800 x me.
Charge = -1 x (e- charge). = +1.602176462 x 10-19 C = +1 atomic units
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How were p+ and e- arranged?
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Thompson:• Ball of uniform positive charge, with small negative
dots (e-) stuck in it.• The “plum-pudding” model.
19101910 Rutherford (former Thompson graduate student) fired α-particles at thin metal foils.Expected them to pass through with minor deflections.
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Rutherford Scattering Experiment
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Different Models of the Atom: different scattering results
α particles
α particles
“Plum pudding model”•+ and – charges evenly distrubted•low, uniform density of matter•No back scattering
Rutherford’s explanation of results:
Small regions of very high density+ charge in the dense regions- Charges in region around it
From wikipedia14
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Some Large Deflections were osbservedSome Large Deflections were osbserved
α particles
Rutherford“It was about as credible as if you had fired a 15-inch shell at a piece of paper and it came back and hit you.”
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≈10,000 times smaller diameter than the entire atom. e- occupy the remaining space.
α particles
Most of the mass and all “+” charge is concentrated in a small core, the nucleusnucleus.
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Nucleus diameter~ 10-4 Å = 10-14 metersMass ~ 10-27 Kg
Charge cloud Diameter ~ 1 ÅMass ~ 10-30 kg
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Most Chemistry involves rearrangement of outermost electrons, not nuclei
Example: H 1p+ , 1 e-
H + H H2
+
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7 Å Epitaxial Al2O3(111) film on Ni3Al(111) (Kelber group):•Grown in UHV•Uniform•No Charging
S. Addepalli, et al. Surf. Sci. 442 (1999) 3464
STM
Start with ordered films growth studiesProceed to amorphous films on Si(100)
Surface terminated by hexagonal array of O anions
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•Atomic mass > mass of all p+ and e- in an atom.•Rutherford proposed a neutral particle.
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mn ≈ mp (0.1% larger).
mn = 1.67492728 x 10-24 g.
Present in all atoms (except ‘normal’ H).
Chadwick (1932) fired -particles at Be atoms. Neutral particles, neutronsneutrons, were ejected:
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~ 10-10 meters = 1 angstrom (Å)
_
+10-14 m
Smeared out electron charge cloud
+++
++
++
Protons and neutrons
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NucleusNucleus• Contains p+ and n0
• Most of the atomic mass.• Small (~10,000x smaller diameter than the atom).• Positive (each p+ has +1 charge).
• Small light particles surrounding the nucleus.• Occupy most of the volume.• Charge = -1.
Atoms are neutral. Number of e− = Number of p+
ElectronsElectrons
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A neutron can decay into a proton and electron:
n0 p+ + e-
This can cause decay of a radioactive element, e.g.,
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6C# of p+ + n0
Atomic No.(# e- = # p+
Elemental symbol (carbon)
Carbon with 6 protons and 8 neutrons is unstable (radioactive)
Carbon with 6 protons and 6 neutrons is stable (non-radioactive
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6C12
6Cradioctive stable
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An atom of 14C can undergo decay to N as a neutron turns into a proton + an emitted electron
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6C14
7 N + e-
1 p+ 1 n0 + an electron (emitted)
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Same element - same number of p+
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Atomic numberAtomic number (Z) = number of p+
1 amu = 1.66054 x 10-24 g
Particle Mass Mass Charge (g) (amu) (atomic units)
e− 9.1093826 x 10-28 0.000548579 −1
p+ 1.67262129 x 10-24 1.00728 +1
n0 1.67492728 x 10-24 1.00866 0
Atomic mass unit (amu) = (mass of C atom) that contains 6 p+ and 6 n0.
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IsotopesIsotopes Atoms of the same elementsame element with different A.• equal numbers of p+
• different numbers of n0
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deuterium (D)
tritium (T)
Hydrogen isotopes: H 1 p+, 0 n011
21H 1 p+, 1 n0
31H 1 p+, 2 n0
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ISOTOPES: SAME Element, Different numbers of neutrons
C12 C14
Carbon: atomic no. = 6 6 protons in the nucleus+ 6 electrons
Atomic mass = 12 amu (12 gr/mole)
Therefore , 6 protons + 6 neutrons
Atomic mass = 14 amu
Therefore, 6 protons+ 8 neutrons
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Isotopes Display the same chemical reactivities (which depend mainly on the outer arrangement of the electrons)
12C + O2 CO2
14C +O2 CO2
Isotopes display different nuclear properties
C12stable
C14 Radioactive: spontaneously emits electrons. Half-life ~ 5730 years
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Isotopes and Moles (more on this later) and isotope abundance:
1 mole = 6.02 x 1023 of anything!
1 mole of atoms = 6.02 x 1023 atoms
Molar Mass (in grams) = average atomic mass (in amu)
1 mole of H atoms = 1.008 gr.
Why not 1.000 gr?? most atoms are 1H, but some are 2H (deuterium)
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Average atomic mass of H = 1.008 amu
100 atoms have a mass of 100.8 amu
# of 2H atoms = n# of 1H atoms = 100 –n (assume these are the only two isotopes that matter)
Mass of 100 atoms = n x 2.000 +(100-n) x 1.000 = 100.8 amu
2n + 100-n = 100.8
n = 0.8 So, out of every 100 atoms , have 0.8 2H atoms Out of every 1000 atoms, have 8 2H atoms
Natural abundance of “heavy hydrogen (deuterium) is then 0.8%
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Most elements occur as a mixture of isotopes.
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Magnesium is a mixture of:
24Mg 25Mg 26Mg
number of p+ 12 12 12
number of n0 12 13 14
mass / amu 23.985 24.986 25.982
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For most elements, the percent abundancepercent abundance of its isotopes are constant (everywhere on earth).
The periodic table lists an average atomic weight.
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ExampleExample
Boron occurs as a mixture of 2 isotopes, 10B and 11B. The abundance of 10B is 19.91%. Calculate the atomic weight of boron.
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Atomic weight for B = 1.994 + 8.817 amu = 10.811 amu
Atomic mass = Σ(fractional abundance)(isotope mass)
(11.0093 amu) = 8.817 amu11B 80.09100
(10.0129 amu)10B 19.91100
= 1.994 amu
% abundance of 11B = 100% - 19.91% = 80.09%
Boron occurs as a mixture of 2 isotopes, 10B and 11B. The abundance of 10B is 19.91%. Calculate the atomic weight of boron.
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B10.811
Boron
5 Atomic number (Z)
Symbol
Name
Atomic weightAtomic weight
Periodic table:
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A counting unit – a familiar counting unit is a “dozen”:
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1 dozen eggs = 12 eggs
1 dozen donuts = 12 donuts
1 dozen apples = 12 apples
1 mole (mol) = Number of atoms in 12 g of 12C• Latin for “heap” or “pile”• 1 mol = 6.02214199 x 1023 “units”• Avogadro’s numberAvogadro’s number
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A green pea has a ¼-inch diameter. 48 peas/foot.
(48)3 / ft3 ≈ 1 x 105 peas/ft3. V of 1 mol ≈ (6.0 x 1023 peas)/(1x 105 peas/ft3) ≈ 6.0 x 1018 ft3
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height = V / area, 1 mol would cover the U.S. to:
U.S. surface area = 3.0 x 106 mi2
= 8.4 x 1013 ft2
6.0 x 1018 ft3
8.4 x 1013 ft2=7.1 x 104 ft = 14 miles !
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1 mole of an atom = atomic weight in grams.
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1 Xe atom has mass = 131.29 amu
1 mol of Xe atoms has mass = 131.29 g
There are 6.022 x 1023 atoms in 1 mol of He andand 1 mol of Xe – but they have different masses.
1 He atom has mass = 4.0026 amu
1 mol of He has mass = 4.0026 g
… 1 dozen eggs is much heavier than 1 dozen peas!
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ExampleExampleHow many moles of copper are in a 320.0 g sample?
Cu-atom mass = 63.546 g/mol (periodic table)
Conversion factor: 1 mol Cu63.546 g
= 1
nCu = 320.0 g x 1 mol Cu63.546 g = 5.036 mol Cu
n = number of moles
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Calculate the number of atoms in a 1.000 g sample of boron.
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nB = (1.000 g) 1 mol B10.81 g
= 0.092507 mol B
B atoms = (0.092507 mol B)(6.022 1023 atoms/mol)
= 5.571 1022 B atoms
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Dimensional Analysis and Problem SolvingSpecial Homework Problem: Due Tues. Recitation
Density = mass/volume
Problem:
Assume that a hydrogen atom has a spherical diameter of 1 angstrom
Assume that the nucleus (1 proton) has a diameter of 10-4 angstrom
1.Calculate the densities of the nucleus, and of the electron charge cloud in kg/m3
2.Calculate the ratio of the two densities: R = dnucleus/delectron cloud
Mass of proton = 1.67 x 10-27 kg
Mass of electron = 9 x 10-31 kg
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Summarizes• Atomic numbers.• Atomic weights.• Physical state (solid/liquid/gas).• Type (metal/non-metal/metalloid).
Periodicity• Elements with similar properties are
arranged in vertical groups.
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In the USA, “A” denotes a main group element…
…”B” indicates a transition element.
International system uses 1 … 18.
The Periodic Table
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Main group metal
Transition metal
Metalloid
Nonmetal
The Periodic Table
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A period is a horizontal row
Period number
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A group is a vertical column Group 7AHalogens
Group 8ANoble gases
Group 2AAlkaline
earth metals
Group 1AAlkali metals (not H)
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Alkali metals (group 1A; 1)Alkaline earth metals (group 2A; 2)
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• Grey … silvery white colored.
• Highly reactive.• Never found as native
metals.• Form alkaline solutions.
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Transition Elements (groups 1B – 8B)
• Also called transition metals.• Middle of table, periods 4 – 7.• Includes the lanthanides & actinides.
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Lanthanides and Actinides• Listed separately at the bottom.• Chemically very similar.• Relatively rare on earth.
(old name: rare earth elements)
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Groups 3A to 6AGroups 3A to 6A• Most abundant elements in the Earth’s crust
and atmosphere.• Most important elements for living organisms.
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Halogens (group 7A; 17)Halogens (group 7A; 17)• Very reactive non metals.• Form salts with metals.• Colored elements.
Noble gases (8A; 18)Noble gases (8A; 18)• Very low reactivity.• Colorless, odorless gases.
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Atoms are very small.• 1 tsp of water contains 3x as many atoms as
there are tsp of water in the Atlantic Ocean!
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Impractical to use pounds and inches...
Need a universal unit system• The metric system.• The SI system (Systeme International) - derived
from the metric system.
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• A decimal system.• Prefixes multiply or divide a unit by multiples of ten.
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Prefix Factor Example
mega M 106 1 megaton = 1 x 106 tons
kilo k 103 1 kilometer (km) = 1 x 103 meter (m)
deci d 10-1 1 deciliter (dL) = 1 x 10-1 liter (L)
centi c 10-2 1 centimeter (cm) = 1 x 10-2 m
milli m 10-3 1 milligram (mg) = 1 x 10-3 gram (g)
micro μ 10-6 1 micrometer (μm) = 1 x 10-6 m
nano n 10-9 1 nanogram (ng) = 1 x 10-9 g
pico p 10-12 1 picometer (pm) = 1 x 10-12 m
femto f 10-15 1 femtogram (fg) = 1 x 10-15 g
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1 pm = 1 x 10-12 m ; 1 cm = 1 x 10-2 m
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How many copper atoms lie across the diameter of a penny? A penny has a diameter of 1.90 cm, and a copper atom has a diameter of 256 pm.
x 1 x 10-2 m 1 cm
= 7.42 x 107 Cu atoms
1 pm1 x 10-12 m
x1.90 cm = 1.90 x 1010 pm
Number of atoms across the diameter:
1.90 x 1010 pm x 1 Cu atom 256 pm
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LengthLength 1 kilometer = 0.62137 mile1 inch = 2.54 cm (exactly)1 angstrom (Å) = 1 x 10-10 m
VolumeVolume 1 liter (L) = 1000 cm3 = 1000 mL= 1.056710 quarts
1 gallon = 4 quarts = 8 pints
MassMass 1 amu = 1.66054 x 10-24 g1 pound = 453.59237 g = 16 ounces1 ton (metric) = 1000 kg1 ton (US) = 2000 pounds
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Example: What is the volume of a 2 gallon container in Liters?
1 gallon x 4 quarts/gallon = 4 quarts
4 quarts x 1Liter/1.057 quarts = 3.784 Liters (L)
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165 mg dL
A patient’s blood cholesterol level measured 165 mg/dL. Express this value in g/L
1 mg = 1 x 10-3 g ; 1 dL = 1 x 10-1 L
x 1 x10-3 g 1 mg
= 1.65 g/Lx 1 dL1 x10-1 L
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All measurements involve some uncertainty.
Reported numbers include oneone uncertain digit.
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Consider a reported mass of 6.3492 g• Last digit (“2”) is uncertain• Close to 2, but may be 4, 1, 0 …• Five significant figuressignificant figures in this number.
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Read numbers from left to right.Count all digits, startingstarting with the 1st non-zero digit.
AllAll digits are significant exceptexcept zeros used to position a decimal point (“placeholders”).
0.00024030 5 sig. figs. (2.4030 x 10-4)
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placeholders significant
significant
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Round 37.663147 to 3 significant figures.
Examine the 11stst non-significant digit non-significant digit. If it:
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• > 5, round up.• < 5, round down.• = 5, check the 2nd non-significant digit.
round up if absent or odd; round down if even.
last retained digit
1st non-significant digit
Rounds up to 37.72nd non-
significant digit